On n-Trivial Extensions of Rings

arXiv:1604.01486v1 [math.RA] 6 Apr 2016

On n-trivial Extensions of Rings

D. D. Anderson1 , Driss Bennis2,a , Brahim Fahid2,b and Abdulaziz Shaiea2,c 1: Department of Mathematics, The University of Iowa, Iowa City, IA 52242-1419, USA. [email protected] 2: Department of Mathematics, Faculty of Sciences, B.P. 1014, Mohammed V University, Rabat, Morocco. a: [email protected]; driss [email protected] b: [email protected] c: [email protected]

Abstract.

The notion of trivial extension of a ring by a module has been extensively studied and used in ring theory as well as in various other areas of research like cohomology theory, representation theory, category theory and homological algebra. In this paper we extend this classical ring construction by associating a ring to a ring R and a family M = (Mi )ni=1 of n R-modules for a given integer n ≥ 1. We call this new ring construction an n-trivial extension of R by M . In particular, the classical trivial extension will be just the 1-trivial extension. Thus we generalize several known results on the classical trivial extension to the setting of n-trivial extensions and we give some new ones. Various ring-theoretic constructions and properties of n-trivial extensions are studied and a detailed investigation of the graded aspect of n-trivial extensions is also given. We end the paper with an investigation of various divisibily properties of n-trivial extensions. In this context several open questions arise. 2010 Mathematics Subject Classification. primary 13A02, 13A05, 13A15, 13B99, 13E05, 13F05, 13F30; secondary 16S99, 17A99. Key Words. trivial extension; n-trivial extension; graded rings; homogeneous ideal.

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D. D. Anderson et al.

Introduction

Except for a brief excursion in Section 2, all rings considered in this paper are assumed to be commutative with an identity; in particular, R denotes such a ring, and all modules are assumed to be unitary left modules. Of course left-modules over a commutative ring R are actually Rbimodules with mr := rm. The ring Z/nZ of the residues modulo an integer n ∈ N will be noted by Zn . The set N ∪ {0} will be denoted by N0 . Recall that the trivial extension of R by an R-module M is the ring denoted by R ⋉ M whose underlying additive group is R⊕M with multiplication given by (r, m)(r ′ , m′ ) = (rr ′ , rm′ +mr ′ ). Since its introduction by Nagata in [40], the trivial extension of rings (also called idealization since it reduces questions about modules to ideals) has been used by many authors and in various contexts in order to produce examples of rings satisfying preassigned conditions (see, for instance, [9] and [38]). It is known that the trivial extension R ⋉ M is related to the following two ring constructions (see for instance [9, Section 2]): Generalized triangular matrix ring. Let R := (Ri )ni=1 be a family of rings and M := (Mi,j )1≤i n. b

a+b

a,b

b a,b

a,b

Then R ⋉n M1 ⋉ · · · ⋉ Mn is a Zn+1 -graded ring R0 ⊕ R1 ⊕ · · · ⊕ Rn where R0 = R and Ra = Ma for a = 1, ..., n.

As a Γn+1 -graded ring. Here Γn+1 = {0, 1, ..., n} is a commutative monoid with addition b := i + j if i + j ≤ n and i+j b := 0 if i + j > n (so Z2 and Γ2 are isomorphic). In this i+j b 6= 0 and case, we define maps ϕ bi,j , for i, j ∈ Γn+1 , by ϕ bi,j = ϕi,j when i = j = 0 or i+j b = 0. Then R ⋉n M1 ⋉ · · · ⋉ Mn is a ϕ bi,j : Mi × Mj −→ M0 = R to be the zero map when i+j Γn+1 -graded ring R0 ⊕ R1 ⊕ · · · ⊕ Rn where R0 = R and Ri = Mi for 1 ≤ i ≤ n. Note that each of these gradings have the same set of homogeneous elements. ∞

We have observed that R ⋉n M1 ⋉ · · · ⋉ Mn is an N0 -graded ring ⊕ Ri where R0 = R, Ri = Mi i=0

for i = 1, ..., n and Ri = 0 for i > n. So R ⋉n M1 ⋉ · · · ⋉ Mn is a graded ring isomorphic to ∞ ⊕ Ri / ⊕ Ri . The following result presents the converse implication. Namely, it shows that i=0

i≥n+1

the n-trivial extensions can be realised as quotients of graded rings. ∞

Proposition 3.1 Let ⊕ Si be an N0 -graded ring and m ∈ N. Then S0 ⋉m S1 ⋉ · · · ⋉ Sm with i=0 ∞

∞

i=0

i=0

the product induced by ⊕ Si is naturally an N0 -graded ring isomorphic to ⊕ Si /

⊕

i≥m+1

Si .

Proof. Obvious. The following result presents a particular case of Proposition 3.1. Proposition 3.2 For an R-module N , we have the following two natural ring isomorphisms: TR (N )/ SR (N )/

⊕ TRi (N ) ∼ = R ⋉n N ⋉ TR2 (N ) ⋉ · · · ⋉ TRn (N ), and

i≥n+1

i (N ) ∼ R ⋉ N ⋉ S 2 (N ) ⋉ · · · ⋉ S n (N ). ⊕ SR = n R R

i≥n+1

2 (N )⋉· · ·⋉S n (N ) Moreover, suppose that N is a free R-module with a basis B, then R⋉n N ⋉SR R n+1 is (graded) isomorphic to R[{Xb }b∈B ]/({Xb }b∈B ) . In particular, R ⋉n R ⋉ · · · ⋉ R with the natural maps is isomorphic to R[X]/(X n+1 ).

Proof. Obvious.

n-Trivial Extensions of Rings

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Our next result shows that the n-trivial extension of a graded ring by graded modules has a natural grading. It is an extension of [9, Theorem 4.5]. Theorem 3.3 Let Γ be a commutative additive monoid. Assume that R = ⊕ Ri is Γ-graded α∈Γ

and Mi = ⊕ Mαi is Γ-graded as an R-module for every i ∈ {1, ..., n}, such that ϕi,j (Mαi , Mβj ) ⊆ α∈Γ

i+j Mα+β . Then R ⋉n M1 ⋉ · · · ⋉ Mn is a Γ-graded ring with (R ⋉n M1 ⋉ · · · ⋉ Mn )α = Rα ⊕ Mα1 ⊕ · · · ⊕ Mαn .

Proof. Similar to the proof of [9, Theorem 4.5]. In the case where R is either a polynomial ring or a Laurent polynomial ring we get the following result in which the first assertion is an extension of [9, Corollary 4.6 (1)]. Corollary 3.4 The following statements are true. 1. (R ⋉n M1 ⋉ · · · ⋉ Mn )[{Xα }] ∼ = R[{Xα }] ⋉n M1 [{Xα }] ⋉ · · · ⋉ Mn [{Xα }] for any set of indeterminates {Xα } over R.

2. (R ⋉n M1 ⋉ · · · ⋉ Mn )[{Xα±1 }] ∼ = R[{Xα±1 }] ⋉n M1 [{Xα±1 }] ⋉ · · · ⋉ Mn [{Xα±1 }] for any set of indeterminates {Xα } over R.

Also, as in the classical case, we get the related (but not graded) power series case. It is a generalization of [9, Corollary 4.6 (2)]. First recall that, for a given set of analytic indeterminates {Xα }α∈Λ over R, we can consider three types of power series rings (see [44] for further details about generalized power series rings): R[[{Xα }α∈Λ ]]1 ⊆ R[[{Xα }α∈Λ ]]2 ⊆ R[[{Xα }α∈Λ ]]3 . Here R[[{Xα }α∈Λ ]]1 = ∪{R[[{X α1 , ..., Xαn }]]|{α1 , ..., αn } ⊆ Λ}, P R[[{Xα }α∈Λ ]]2 = {P∞ f i=0 i |fi ∈ R[{Xα }α∈Λ ] is homogeneous of degree i} and ∞ R[[{Xα }α∈Λ ]]3 = { i=0 fi |fi is a possibly inf inite sum of monomials of degree i with at most one monomial of the f orm rα1 ,...,αn Xαi11 · · · Xαinn f or each set {α1 , ..., αn } with i1 + · · · + in = i}. More generally, given a partially ordered additive (S, +, ≤), the generalized power P monoid s ≤ as X where supp(f ) is Artinian and narrow series ring R[[X, S ]] consists of all formal sums s∈S

(i.e., has no infinite family of incomparable elements) where addition and multiplication are carried out in the usual way. If Λ is a well-ordered set, S = ⊕ N0 and ≤ is the reverse λ∈Λ

lexicographic order on S, then R[[X, S ≤ ]] ∼ = R[[{Xα }]]3 . Note, that in a similar manner we can define three types of power series over a module. The routine proof of the following theorem is left to the reader.

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Theorem 3.5 1, 2, 3,

1. Let {Xα }α∈Λ be a set of analytic indeterminates over R. Then, for i =

(R⋉n ⋉M1 ⋉· · ·⋉Mn )[[{Xα }α∈Λ ]]i ∼ = R[[{Xα }α∈Λ ]]i ⋉n M1 [[{Xα }α∈Λ ]]i ⋉· · ·⋉Mn [[{Xα }α∈Λ ]]i . 2. Let (S, +, ≤) be a partially ordered additive monoid. Then (R ⋉n ⋉M1 ⋉ · · · ⋉ Mn )[[X, S ≤ ]] ∼ = R[[X, S ≤ ]] ⋉n M1 [[X, S ≤ ]] ⋉ · · · ⋉ Mn [[X, S ≤ ]]. Now, we give, as an extension of [9, Theorem 4.1], the following result which investigates the localization of an n-trivial extension. For this we need the following technical lemma. Lemma 3.6 For every (mi ) ∈ R ⋉n M and every k ∈ {1, ..., n}, (m0 , 0, ..., 0, mk , mk+1 , ..., mn )(m0 , 0, ..., 0, −mk , 0, ..., 0) = (m20 , 0, ..., 0, ek+1 , ..., en ) where el = m0 ml − mk ml−k for every l ∈ {k + 1, ..., n}. Consequently, there is an element (fi ) of R ⋉n M , such that n (mi )(fi ) = (m20 , 0, ..., 0). n

We will denote the element (fi ) in Lemma 3.6 by (f mi ) so (mi )(f mi ) = (m20 , 0, ..., 0). Theorem 3.7 Let S be a multiplicatively closed subset of R and N = (Ni ) be a family of Rmodules where Ni is a submodule of Mi for each i ∈ {1, ..., n} and Ni Nj ⊆ Ni+j for every 1 ≤ i, j ≤ n − 1 and i + j ≤ n. Then the set S ⋉n N is a multiplicatively closed subset of R ⋉n M and we have a ring isomorphism (R ⋉n M )S⋉n N ∼ = RS ⋉n MS where MS = (MiS ). Proof. It is trivial to show that S ⋉n N is a multiplicatively closed subset of R ⋉n M . Now in order to show the desired isomorphism, we need to make, as done in the proof of [9, Theorem 4.1 (1)], the following observation: Let (mi ) ∈ R ⋉n M and (si ) ∈ S ⋉n N . Then using the notation of Lemma 3.6, (mi )(sei ) (m′i ) (mi ) = = (si ) (S0 , 0, ..., 0) (S0 , 0, ..., 0) n

where (m′i ) = (mi )(sei ) and S0 = s20 . Then the map f : (R ⋉n M )S⋉n N (mi ) (si )

is the desired isomorphism.

−→ RS ⋉n MS ′ m′ m′ n 7−→ ( S00 , S01 , ..., m S0 )


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As a simple but important particular case of Theorem 3.7, we get the following result which extends [9, Theorem 4.1 and Corollary 4.7]. In Theorem 4.7, we will show that if P is a prime ideal of R, then P ⋉n M is a prime ideal of R ⋉n M . This fact is used in the next result to show that the localization of an n-trivial extension at a prime ideal is isomorphic to an n-trivial extension. In what follows, we use T (A) to denote the total quotient ring of a ring A. In Proposition 4.9, we will prove that S ⋉n M , where S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )), is the set of all regular elements of R ⋉n M . Thus T (R ⋉n M ) = (R ⋉n M )S⋉n M . Corollary 3.8 The following assertions are true. 1. Let P be a prime ideal of R. Then we have a ring isomorphism (R ⋉n M )P ⋉n M ∼ = RP ⋉n MP where MP = (MiP ). 2. We have a ring isomorphism T (R ⋉n M ) ∼ = RS ⋉n MS where S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )). 3. For an indeterminate X over R, we have a ring isomorphism (R ⋉n M1 ⋉ · · · ⋉ Mn )(X) ∼ = R(X) ⋉n M1 (X) ⋉ · · · ⋉ Mn (X). Proof. All the proofs are similar to the corresponding ones for the classical case. Our next result generalizes [9, Theorem 4.4]. It shows that the n-trivial extension of a finite direct product of rings is a finite direct product of n-trivial extensions. For the reader’s convenience we recall here some known facts on the structure of modules over a finite direct product s Y Ri be a finite direct product of rings where s ∈ N. For j ∈ {1, ..., s}, we set of rings. Let R = i=1

R¯j := 0 × · · · × 0 × Rj × 0 × · · · × 0 and, for an R-module N , Nj := R¯j N . Then Nj is a submodule of N and we have N = N1 ⊕ · · · ⊕ Ns . Namely, every element x in N can be written in the form x = x1 + · · · + xs where xj = ej x ∈ Nj for every j ∈ {1, ..., s} (here ej = (0, ..., 0, 1, 0, ..., 0) with 1 in the j’th place). Note that each Nj is also an Rj -module and N1 × · · · × Ns is an R-module ismorphic to N via the following R-isomorphism: −→ N1 × · · · × Ns P N ej x = x 7−→ (e1 x1 , ..., es xs )

and

N1 × · · · × Ns −→ P N (y1 , ..., ys ) 7−→ yj

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Now, consider the family of commutative product maps ϕ = {ϕi,j }

i+j≤n

and define the

1≤i,j≤n−1

following maps: ϕj,i,k : Mj,i × Mj,k −→ Mj,i+k (mj,i , mj,k ) 7−→ ϕj,i,k (mj,i , mj,k ) = ej ϕi,k (mj,i , mj,k ) where Mj,i := R¯j Mi for j ∈ {1, ..., s} and i ∈ {1, ..., n}. It is easily checked that, for every j ∈ {1, ..., s}, ϕj = {ϕj,i,k } i+k≤n is a family of commutative product maps and Rj ⋉ϕj Mj,1 ⋉ 1≤i,k≤n

· · · ⋉ Mj,n is a n-ϕj -trivial extension. Furthermore, ϕi,k : Mi × Mk −→ Mi+k (mi , mk )

7−→ ϕi,k (mi , mk ) =

s P

ϕj,i,k (mj,i , mj,k ).

j=1

With this notation in mind, we are ready to give the desired result. Theorem 3.9 Let R =

s Y i=1

Ri be a finite direct product of rings where s ∈ N. Then

R ⋉ϕ M1 ⋉ · · · ⋉ Mn ∼ = (R1 ⋉ϕ1 M1,1 ⋉ · · · ⋉ M1,n ) × · · · × (Rs ⋉ϕs Ms,1 ⋉ · · · ⋉ Ms,n ). Proof. It is easily checked that the map (r, m1 , ..., mn ) 7−→ ((rj , mj,1 , ..., mj,n ))1≤j≤s is an isomorphism. We end this section with two results which investigate the inverse limit and direct limit of a system of n-trivial extensions. Namely, we show that, under some conditions, the inverse limit or direct limit of a system of n-trivial extensions is isomorphic to an n-trivial extension. The inverse limit case is a generalization of [9, Theorem 4.11]. Let Γ be a directed set and {Mα ; fαβ } be an inverse system of abelian groups over Γ (so for α ≤ β, fαβ : Mβ → Mα ). We know that the inverse limit limMα is isomorphic to the following ←− Q subset of the direct product Mα : α

M∞ := {(xα )α∈Γ |λ ≤ µ ⇒ xλ = fλµ (xµ )}. In the next result, by limMα we mean exactly the set M∞ . ←−

Theorem 3.10 Let Γ be a directed set and n ≥ 1 be an integer. Consider a family of inverse systems {Mi,α ; fi,α,β } over Γ (for i ∈ {0, ..., n}) which satisfy the following conditions: 1. For every α ∈ Γ, M0,α = Rα is a ring,


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2. For every α ∈ Γ and every i ∈ {1, ..., n}, Mi,α is an Rα -module, and 3. For every α ∈ Γ, Rα ⋉n M1,α ⋉ · · · ⋉ Mn,α is an n-trivial extension with a family of commutative product maps: ϕi,j,α : Mi,α × Mj,α −→ Mi+j,α which satisfy, for every α ≤ β, ϕi,j,α (fi,α,β (mi,β ), fj,α,β (mj,β )) = fi+j,α,β (ϕi,j,β (mi,β , mj,β )). Then limRα ⋉n limM1,α ⋉ · · · ⋉ limMn,α is an n-trivial extension with the following family of ←− ←− ←− well-defined commutative product maps: ϕi,j,α : limMi,α × limMj,α → ←−

←−

((mi,α )α , (mj,α )α )

limMi+j,α ←−

7→ (ϕi,j,α (mi,α , mj,α ))α .

Moreover, there is a natural ring ismorphism: lim(Rα ⋉n M1,α ⋉ · · · ⋉ Mn,α ) ∼ = limRα ⋉n limM1,α ⋉ · · · ⋉ limMn,α . ←−

←−

←−

←−

Proof. The result follows using a standard argument. Let Γ be a directed set and {Mγ ; fγλ } a direct system of abelian groups over Γ (so for γ ≤ λ, fγλ : Mγ → Mλ ). We know that the direct limit lim Mγ is isomorphic to ⊕Mγ /S where S is −→ γ generated by all elements λβ (fαβ (aα )) − λα (aα ) where α ≤ β and λλ : Mλ −→ ⊕Mγ is the γ

natural inclusion map for λ ∈ Γ. Since Γ is directed, every element of ⊕Mγ /S has the form γ

λα (aα ) + S for some α ∈ Γ and aα ∈ Mα . Theorem 3.11 Let Γ be a directed set and n ≥ 1 be an integer. Consider a family of direct systems {Mi,α ; fi,α,β } over Γ (for i ∈ {0, ..., n}) which satisfy the following conditions: 1. For every α ∈ Γ, M0,α = Rα is a ring, 2. For every α ∈ Γ and every i ∈ {1, ..., n}, Mi,α is an Rα -module, and 3. For every α ∈ Γ, Rα ⋉n M1,α ⋉ · · · ⋉ Mn,α is an n-trivial extension with a family of commutative product maps: ϕi,j,α : Mi,α × Mj,α −→ Mi+j,α which satisfy, for every β ≤ α, ϕi,j,α (fi,β,α(mi,β ), fj,β,α (mj,β )) = fi+j,β,α(ϕi,j,β (mi,β , mj,β )).

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Then limRα ⋉n limM1,α ⋉ · · · ⋉ limMn,α is an n-trivial extension with the following family of −→ −→ −→ well-defined commutative product maps: ϕi,j,α : limMi,α × limMj,α → −→

−→

((mi,α )α , (mj,α )α )

limMi+j,α −→

7→ (ϕi,j,α (mi,α , mj,α ))α .

Moreover, there is a natural ring ismorphism: lim(Rα ⋉n M1,α ⋉ · · · ⋉ Mn,α ) ∼ = limRα ⋉n limM1,α ⋉ · · · ⋉ limMn,α . −→

−→

−→

−→

Proof. The result follows using a standard argument.

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Some basic algebraic properties of R ⋉n M

In this section we give some basic properties of n-trivial extensions. Before giving the first result, we make the following observations on situations where a subfamily of M is trivial. Observation 4.1 1. If there is an integer i ∈ {1, ..., n − 1} such that Mj = 0 for every j ∈ {i + 1, ..., n}, then there is a natural ring isomorphism R ⋉n M1 ⋉ · · · ⋉ Mi ⋉ 0 ⋉ · · · ⋉ 0 ∼ = R ⋉i M1 ⋉ · · · ⋉ Mi .

If M1 = · · · = Mn−1 = 0, then R ⋉n M can be represented as R ⋉1 Mn . However, if n ≥ 3 and there is an integer i ∈ {1, ..., n − 2} such that, for j ∈ {1, ..., n}, Mj = 0 if and only if j ∈ {1, ..., i}, then R ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mi+1 ⋉ · · · ⋉ Mn cannot be represented as an n − i-trivial extension since (when, for example, 2i + 2 ≤ n) ϕi+1,i+1 (Mi+1 , Mi+1 ) is a subset of M2i+2 not of Mi+2 . 2. If M2k = 0 for every k ∈ N with 1 ≤ 2k ≤ n, then R ⋉n M can be represented as the trivial extension of R by the R-module M1 × M3 × · · · × M2n′ +1 where 2n′ + 1 is the biggest odd integer in {1, ..., n}. Namely, there is a natural ring isomorphism R ⋉n M ∼ = R ⋉1 (M1 × M3 × · · · × M2n′ +1 ).

3. If M2k+1 = 0 for every k ∈ N with 1 ≤ 2k+1 ≤ n, then there is a natural ring isomorphism R ⋉n M ∼ = R ⋉n′′ M2 ⋉ M4 ⋉ · · · ⋉ M2n′′

where 2n′′ is the biggest even integer in {1, ..., n}. In general, for every cyclic submonoid G of Γn+1 generated by an element g ∈ {1, ..., n}, if Mi = 0 if and only if i 6∈ G, then there is a natural ring isomorphism R ⋉n M ∼ = R ⋉s Mg ⋉ M2g ⋉ · · · ⋉ Msg where sg is the biggest integer in G ∩ {1, ..., n}.


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As observed above, if one would discuss according to whether a subfamily of M is trivial or not, then various situations may occur. Thus, for the sake of simplicity, we make the following convention. Convention 4.2 Unless explicitly stated otherwise, when we consider an n-trivial extension for a given n, then we implicitly suppose that Mi 6= 0 for every i ∈ {1, ..., n}. This will be used in the sequel without explicit mention. Note also that the nature of the maps ϕi,j can affect the structure of the n-trivial extension. For example, in case where n = 2, if ϕ1,1 = 0, then R ⋉2 M1 ⋉ M2 ∼ = R ⋉ (M1 × M2 ). For ∼ example, if I ⊆ J is an extension of ideals of R, then R ⋉2 I ⋉ R/J = R ⋉ (I × R/J). Let us start with the following result which presents some relations (easily established) between n-trivial extensions. Proposition 4.3 The following assertions are true. 1. Let G be a submonoid of Γn+1 and consider the family of R-modules M ′ = (Mi′ )ni=1 such that Mi′ = Mi if i ∈ G and Mi′ = 0 if i 6∈ G. Then we have the following (natural) ring extensions: R ֒→ R ⋉n M ′ ֒→ R ⋉n M. In particular, for every m ∈ {1, ..., n}, we have the following (natural) ring extensions: R ֒→ R ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mm ⋉ · · · ⋉ Mn ֒→ R ⋉n M1 ⋉ · · · ⋉ Mn . The extension R ֒→ R ⋉n M1 ⋉ · · · ⋉ Mn will be denoted by in . 2. For every m ∈ {1, ..., n}, 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mm ⋉ · · · ⋉ Mn is an ideal of R ⋉n M and an R ⋉j M1 ⋉ · · · ⋉ Mj -module for every j ∈ {n − m, ..., n} via the action (x0 , x1 , ..., xj )(0, ..., 0, ym , ..., yn ) := (x0 , x1 , ..., xj , 0, ..., 0)(0, ..., 0, ym , ..., yn ) = (x0 , x1 , ..., xn−m , 0, ..., 0)(0, ..., 0, ym , ..., yn ). Moreover, the structure of 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mm ⋉ · · · ⋉ Mn as an ideal of R ⋉n M is the same as the R ⋉j M1 ⋉ · · · ⋉ Mj -module structure for every j ∈ {n − m, ..., n}. In particular, the structure of the ideal 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn is the same as the one of the R-module Mn . 3. For every m ∈ {1, ..., n}, we have the following natural ring isomorphisms: R ⋉n M1 ⋉ · · · ⋉ Mn /0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mm ⋉ · · · ⋉ Mn ∼ = R ⋉m−1 M1 ⋉ · · · ⋉ Mm−1

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D. D. Anderson et al. obtained from the natural ring homomorphism: πm−1 : R ⋉n M1 ⋉ · · · ⋉ Mn −→ R ⋉m−1 M1 ⋉ · · · ⋉ Mm−1 (r, x1 , ..., xn ) 7−→ (r, x1 , ..., xm−1 ) where for m = 1, R ⋉m−1 M1 ⋉ · · · ⋉ Mm−1 = R.

To give another example for the assertion (1), one can show that, for n = 3, {0, 2} is a submonoid of Γ4 . Then we have the following (natural) ring extensions: R ֒→ R ⋉ M2 ֒→ R ⋉3 M1 ⋉ M2 ⋉ M3 . Remark 4.4 We have seen that, in the case of n = 1, the ideal structure of 0 ⋉1 M1 is the same as the R-module structure of 0 ⋉1 M1 . Actually, Nagata [40] used this to reduce proofs of module-theoretic results to the ideal case. However, for n ≥ 2, the R-module structure of 0 ⋉n M1 ⋉ · · · ⋉ Mn need not be the same as the ideal structure. For instance, consider the 2trivial extension Z⋉2 Z⋉Z (with the maps induced by the multiplication in Z). Then Z(0, 1, 1) = {(0, m, m)|m ∈ Z} while the ideal of Z ⋉2 Z ⋉ Z generated by (0, 1, 1) is 0 ⋉2 Z ⋉ Z. However, according to Proposition 4.3 (2), (Z ⋉1 Z)(0, 1, 1) = (Z ⋉2 Z ⋉ Z)(0, 1, 1). The notion of extensions of ideals under ring homomorphisms is a natural way to construct examples of ideals. In this context, we use the ring homomorphism im (indicated in Proposition 4.3) to give such examples. Proposition 4.5 For an ideal I of R, we have the following assertions: 1. The ideal I ⋉n IM1 ⋉ · · · ⋉ IMn of R ⋉n M is the extension of I under the ring homomorphism in , and we have the following natural ring isomorphism: (R ⋉n M )/(I ⋉n IM1 ⋉ · · · ⋉ IMn ) ∼ = (R/I) ⋉n (M1 /IM1 ) ⋉ · · · ⋉ (Mn /IMn ) where the multiplications are well-defined as follows: ϕi,j : Mi /IMi × Mj /IMj (mi , mj )

−→ Mi+j /IMi+j 7−→ mi mj := ϕi,j (mi , mj ) := ϕi,j (mi , mj ) = mi mj .

2. The ideal I ⋉n IM1 ⋉ · · · ⋉ IMn is finitely generated if and only if I is finitely generated. Proof. 1. The proof is straightforward. 2. Using π0 it is clear that if I ⋉n IM1 ⋉· · ·⋉IMn is generated by elements (rj , mj,1 , ..., mj,n ) with j ∈ E for some set E, then I is generated by the rj ’s. Conversely, if I is generated by elements rj with j ∈ E for some set E, then I ⋉n IM1 ⋉ · · · ⋉ IMn is generated by the (rj , 0, ..., 0)’s.


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Now, we determine the radical, prime and maximal ideals of R ⋉n M . As in the classical case, we show that these ideals are particular cases of the homogenous ones, which are characterized in the next section. However, we give these particular cases here because of their simplicity which is reflected, using the following lemma, on the fact that they contain the nilpotent ideal 0 ⋉n M (of index n + 1). Lemma 4.6 Every ideal of R ⋉n M which contains 0 ⋉n M has the form I ⋉n M for some ideal I of R. In this case, we have the following natural ring isomorphism: R ⋉n M/I ⋉n M ∼ = R/I. Proof. Let J be an ideal of R ⋉n M which contains 0 ⋉n M and consider the ideal I = π0 (J) of R where π0 is the surjective ring homomorphism used in Proposition 4.3. Then J ⊆ I ⋉n M and by the fact that 0 ⋉n M ⊆ J, we deduce that J = I ⋉n M . Finally, using π0 and the fact that π0−1 (I) = J, we get the desired isomorphism. The following result is an extension of [9, Theorem 3.2]. Theorem 4.7 Radical ideals of R ⋉n M have the form I ⋉n M where I is a radical ideal of R. In particular, the maximal (resp., the prime) ideals of R ⋉n M have the form M ⋉n M (resp, P ⋉n M ) where M (resp., P ) is a maximal (resp., a prime) ideal of R. Proof. Using Lemma 4.6, it is sufficient to note that every radical ideal contains 0 ⋉n M since (0 ⋉n M )n+1 = 0. Theorem 4.7 allows us to easily determine both the Jacobson radical and the nilradical of R ⋉n M . Corollary 4.8 The Jacobson radical J(R ⋉n M ) (resp., the nilradical Nil(R ⋉n M )) of R ⋉n M is J(R) ⋉n M (resp., Nil(R) ⋉n M ) and the Krull dimension of R ⋉n M is equal to that of R. We end this section with an extension of [9, Theorems 3.5 and 3.7] which determines, respectively, the set of zero divisors Z(R ⋉n M ), the set of units U (R ⋉n M ) and the set of idempotents Id(R ⋉n M ) of R ⋉n M . It is worth noting that trivial extensions have been used to construct examples of rings with zero divisors that satisfies certain properties. As mentioned in the introduction, particular 2-trivial extensions are used to settle some questions. Recently, in [14], a 2-trivial extension is used in the context of zero-divisor graphs to give an appropriate example. Proposition 4.9 The following assertions are true. 1. The set of zero divisors of R ⋉n M is Z(R⋉n M ) = {(r, m1 , ..., mn )|r ∈ Z(R)∪Z(M1 )∪· · ·∪Z(Mn ), mi ∈ Mi f or i ∈ {1, ..., n}}.

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D. D. Anderson et al. Hence S ⋉n M where S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )) is the set of regular elements of R ⋉n M . 2. The set of units of R ⋉n M is U (R ⋉n M ) = U (R) ⋉n M . 3. The set of idempotents of R ⋉n M is Id(R ⋉n M ) = Id(R) ⋉n 0.

Proof. All the proofs are similar to the corresponding ones for the classical case. For completeness, we give a proof of the first assertion. Let (r, m1 , ..., mn ) ∈ R ⋉n M such that r ∈ Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn ). If r = 0, then (0, m1 , ..., mn )(0, ..., m′n ) = (0, ..., 0) for every m′n ∈ Mn . Hence (r, m1 , ..., mn ) ∈ Z(R ⋉n M ). Suppose r 6= 0. If r ∈ Z(R), there exists a nonzero element s ∈ R such that rs = 0, so (r, 0, ..., 0)(s, 0, ..., 0) = (0, ..., 0) and hence (r, 0, ..., 0) ∈ Z(R ⋉n M ). If r ∈ Z(Mi ), for some i ∈ {1, ..., n}, there exists a nonzero element m′′i of Mi such that rm′′i = 0, so (r, 0, ..., 0)(0, ..., 0, m′′i , 0, ..., 0) = (0, ..., 0). Hence (r, 0, ..., 0) ∈ Z(R⋉n M ). Now, since Z(R⋉n M ) is a union of prime ideals and Nil(R⋉n M ) is contained in each prime ideal and using the fact that (0, m1 , ..., mn ) ∈ Nil(R ⋉n M ), we conclude that (r, m1 , ..., mn ) = (r, 0, ..., 0) + (0, m1 , ..., mn ) ∈ Z(R ⋉n M ). This gives the first inclusion. Conversely, let (r, m1 , ..., mn ) ∈ Z(R⋉n M ). Then there is (s, m′1 , ..., m′n ) ∈ R⋉n M −{(0, ..., 0)} such ..., 0) = (r, m1 , ..., mn )(s, m′1 , ..., m′n ) = (rs, rm′1 +sm1 , rm′2 +m1 m′1 +sm2 , ..., rm′n + P that (0, mi m′j + smn ). If s 6= 0, then r ∈ Z(R), and if s = 0, we get r ∈ Z(M1 ) if m′1 6= 0, otherwise i+j=n

we pass to m′2 and so on we continue until we arrive at s = 0 and m′i = 0 for all i ∈ {1, ..., n − 1}. Then rm′n = 0 and m′n 6= 0, so r ∈ Z(Mn ). This gives the desired inclusion.

5

Homogeneous ideals of n-trivial extensions

The study of the classical trivial extension as a graded ring established some interesting properties (see, for instance, [9, Section 3]). Namely, in [9], studying homogeneous ideals of the trivial extension shed more light on the structure of their ideals. Then naturally one would like to extend this study to the context of n-trivial extensions. In this section we extend this study to the context of n-trivial extensions, where here R ⋉n M is a (N0 -)graded ring with, as indicated in Section 3, (R ⋉n M )0 = R, (R ⋉n M )i = Mi , for every i ∈ {1, ..., n}, and (R ⋉n M )i = 0 for every i ≥ n + 1. Note that we could also consider R ⋉n M as a Zn+1 -graded ring or Γn+1 -graded ring as mentioned in Section 3. For that, it is convenient to recall the following definitions: Let Γ be a commutative additive monoid and S = ⊕ Sα be a Γ-graded ring. Let N = ⊕ Nα be a Γ-graded S-module. For every α∈Γ

α∈Γ


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α ∈ Γ, the elements of Nα are said to be homogeneous of degree α. A submodule N ′ of N is said to be homogeneous if one of the following equivalent assertions is true. (1) N ′ is generated by homogeneous elements, P (2) If nα ∈ N ′ , where G′ is a finite subset of Γ and each nα is homogeneous of degree α, α∈G′

then nα ∈ N ′ for every α ∈ G′ , or

(3) N ′ = ⊕ (N ′ ∩ Nα ). α∈Γ

In particular, an ideal J of R ⋉n M is homogeneous if and only if J = (J ∩ R) ⊕ (J ∩ M1 ) ⊕ · · · ⊕ (J ∩ Mn ). Note that I := J ∩ R is an ideal of R and, for i ∈ {1, ..., n}, Ni := J ∩ Mi is an R-submodule of Mi which satisfies IMi ⊆ Ni and Ni Mj ⊆ Ni+j for evey i, j ∈ {1, ..., n}. The next result extends [9, Theorem 3.3 (1)]. Namely, it determines the structure of the homogeneous ideals of n-trivial extensions. In what follows, we use the ring homomorphism Π0 := π0 (used in Proposition 4.3) and, for i ∈ {1, ..., n}, the following homomorphism of R-modules: Πi : R ⋉n M1 ⋉ · · · ⋉ Mn −→ Mi (r, m1 , ..., mn ) 7−→ mi . Theorem 5.1 The following assertions are true. 1. Let I be an ideal of R and let C = (Ci )i∈{1,...,n} be a family of R-modules such that Ci ⊆ Mi for every i ∈ {1, ..., n}. Then I ⋉n C is a (homogeneous) ideal of R ⋉n M if and only if IMi ⊆ Ci and Ci Mj ⊆ Ci+j for all i, j ∈ {1, ..., n} with i + j ≤ n. Thus if I ⋉n C is an ideal of R⋉n M , then Mi /Ci is an R/I-module for every i ∈ {1, ..., n}, and we have a natural ring isomorphism (R ⋉n M1 ⋉ · · · ⋉ Mn )/(I ⋉n C1 ⋉ · · · ⋉ Cn ) ∼ = (R/I) ⋉n (M1 /C1 ) ⋉ · · · ⋉ (Mn /Cn ) where the multiplications are well-defined as follows: ϕi,j : Mi /Ci × Mj /Cj (mi , mj )

−→ Mi+j /Ci+j 7−→ mi mj .

In particular, (R ⋉n M1 ⋉ · · · ⋉ Mn )/(0 ⋉n C1 ⋉ · · · ⋉ Cn ) ∼ = R ⋉n (M1 /C1 ) ⋉ · · · ⋉ (Mn /Cn ). 2. Let J be an ideal of R ⋉n M and consider K := Π0 (J) and Ni := Πi (J) for every i ∈ {1, ..., n}. Then,

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D. D. Anderson et al. (a) K is an ideal of R and Ni is a submodule of Mi for every i ∈ {1, ..., n} such that KMi ⊆ Ni and Ni Mj ⊆ Ni+j for every j ∈ {1, ..., n} with i + j ≤ n. Thus K ⋉n N1 ⋉ · · · ⋉ Nn is a homogeneous ideal of R ⋉n M1 ⋉ · · · ⋉ Mn . (b) J ⊆ K ⋉n N1 ⋉ · · · ⋉ Nn . (c) The ideal J is homogeneous if and only if J = K ⋉n N1 ⋉ · · · ⋉ Nn .

Proof. 1. If I ⋉n C1 ⋉· · ·⋉Cn is an ideal of R⋉n M , then (R⋉n M1P ⋉· · ·⋉Mn )(I ⋉n C1 ⋉· · ·⋉Cn ) = I ⋉n (IM1 + C1 ) ⋉ (IM2 + C2 + C1 M1 ) ⋉ · · · ⋉ (IMn + Cn + Ci Mj ). Thus IMi ⊆ Ci and i+j=n

Ci Mj ⊆ Ci+j for every i, j ∈ {1, ..., n}. Conversely, suppose that we have IMi ⊆ Ci and Ci Mj ⊆ Ci+j for all i, j ∈ {1, ..., n} with i + j ≤ n. Then Mi /Ci is an R/I-module for every i ∈ {1, ..., n} and the map f : R ⋉n M1 ⋉ · · · ⋉ Mn −→ (R/I) ⋉n (M1 /C1 ) ⋉ · · · ⋉ (Mn /Cn ) (r, m1 , ..., mn ) 7−→ (r + I, m1 + C1 , ..., mn + Cn ) is a well-defined surjective homomorphism with Kerf = I ⋉n C1 ⋉ · · · ⋉ Cn , so I ⋉n C1 ⋉ · · · ⋉ Cn is an ideal of R ⋉n M1 ⋉ · · · ⋉ Mn and (R ⋉n M1 ⋉ · · · ⋉ Mn )/(I ⋉n C1 ⋉ · · · ⋉ Cn ) ∼ = (R/I) ⋉n (M1 /C1 ) ⋉ · · · ⋉ (Mn /Cn ). In particular, (R ⋉n M1 ⋉ · · · ⋉ Mn )/(0 ⋉n C1 ⋉ · · · ⋉ Cn ) ∼ = R ⋉n (M1 /C1 ) ⋉ · · · ⋉ (Mn /Cn ). 2. All of the three statements are easily checked.

The following result presents some properties of homogeneous ideals of R ⋉n M . It is an extension of both [9, Theorem 3.2 (3)] and [9, Theorem 3.3 (2) and (3)]. In particular, we determine, as an extension of [9, Theorem 3.3 (3)], the form of homogeneous principal ideals. In fact, the characterization of homogeneous principal ideals plays a key role in studying homogeneous ideals. This is due to (the easily checked) fact that an ideal I of a graded ring is homogeneous if every principal ideal generated by an element of I is homogeneous. Proposition 5.2 The following assertions are true. 1. Let I ⋉n N1 ⋉ · · · ⋉ Nn and I ′ ⋉n N1′ ⋉ · · · ⋉ Nn′ be two homogeneous ideals of R ⋉n M . Then we have the following homogeneous ideals of R ⋉n M : (a) (I ⋉n N1 ⋉ · · · ⋉ Nn ) + (I ′ ⋉n N1′ ⋉ · · · ⋉ Nn′ ) = (I + I ′ ) ⋉n (N1 + N1′ ) ⋉ · · · ⋉ (Nn + Nn′ ),

(b) (I ⋉n N1 ⋉ · · · ⋉ Nn ) ∩ (I ′ ⋉n N1′ ⋉ · · · ⋉ Nn′ ) = (I ∩ I ′ ) ⋉n (N1 ∩ N1′ ) ⋉ · · · ⋉ (Nn ∩ Nn′ ),

′ ′ ′ ′ ′ ′ (c) (I ⋉n N1 ⋉ · · · ⋉ Nn )(I ′ ⋉n N1′ ⋉ P· · · ⋉ N′ n ) = II ⋉n (IN1 + I N1 ) ⋉ (IN2 + I N2 + ′ ′ ′ Ni Nj ), and N1 N1 ) ⋉ · · · ⋉ (INn + I Nn + i+j=n


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(d) (I ⋉n N1 ⋉ · · · ⋉ Nn ) : (I ′ ⋉n N1′ ⋉ · · · ⋉ Nn′ ) = ((I :R I ′ ) ∩ (N1 :R N1′ ) ∩ · · · ∩ (Nn :R ′ )) ⋉ · · · ⋉ (Nn :Mn I ′ ) where Nn′ )) ⋉n ((N1 :M1 I ′ ) ∩ (N2 :M1 N1′ ) ∩ · · · ∩ (Nn :M1 Nn−1 (Ni+j :Mi Nj′ ) := {mi ∈ Mi |mi Nj′ ⊆ Ni+j } for every i, j ∈ {0, ..., n} with i + j ≤ n (here M0 = R, N0 = I and N0′ = I ′ ). 2. A principal ideal h(a, m1 , ..., mn )i of R⋉n M is homogeneous if and only if P h(a, m1 , ..., mn )i = aR ⋉n (Rm1 + aM1 ) ⋉ (Rm2 + aM2 + m1 M1 ) ⋉ · · · ⋉ (Rmn + aMn + mi Mj ). i+j=n

√

p

3. For an ideal J of R ⋉n M , J = Π√0 (J) ⋉n M . In particular, √ if I ⋉n C1 ⋉ · · · ⋉ Cn is a homogeneous ideal of R ⋉n M , then I ⋉n C1 ⋉ · · · ⋉ Cn = I ⋉n M . Proof. 1. The proof for each of the first three statements is similar to the corresponding one of [38, Theorem 25.1 (2)]. The last statement easily follows from the fact that the residual of two homogeneous ideals is again homogeneous. 2. Apply assertion (1) and Theorem 5.1 (1). 3. The proof is similar to the one of [9, Theorem 3.2 (3)]. It is a known fact that, in case where n = 1, even if a homogeneous ideal I ⋉ C is finitely generated, the R-module C is not necessarily finitely generated (you can consider Z ⋉ Q and the principal ideal h(2, 0)i = 2Z ⋉ Q as an example). The following result presents, in this context, some particular cases obtained using standard arguments. Proposition 5.3 The following assertions are true. 1. The ideal 0 ⋉n M of R ⋉n M is finitely generated if and only if each R-module Mi is finitely generated. 2. If a homogeneous ideal I ⋉n C1 ⋉ · · · ⋉ Cn of R ⋉n M is finitely generated, then I is a finitely generated ideal of R. The converse implication is true when Ci is a finitely generated R-module for every i ∈ {1, ..., n}. From the previous section, we note that every radical (hence prime) ideal of R ⋉n M is homogeneous. However, it is well-known that the ideals of the classical trivial extensions are not in general homogeneous (see [9]). Then natural questions arise: Question 1: When every ideal of a given class I of ideals of R ⋉n M is homogeneous? Question 2: For a given ring R and a family of R-modules M = (Mi )ni=1 , what is the class of all homogeneous ideals of R ⋉n M ? It is clear that these questions depend on the structure of both R and each Mi . For instance, for n = 1, if R is a quasi-local ring with maximal m, then a proper homogeneous ideal of R⋉R/m

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has either the form I ⋉ R/m or I ⋉ 0 where I is a proper ideal of R. And a proper homogeneous principal ideal of R ⋉ R/m has either the form 0 ⋉ R/m or I ⋉ 0 where I is a principal ideal of R. Then, for instance, a principal ideal of R ⋉ R/m generated by an element (a, e) where a and e are both nonzero with a ∈ m, is not homogeneous.

Question 1 was investigated in [9] for the case where I is the class of regular ideals of R ⋉1 M [9, Theorem 3.9]. Also, under the condition that R is an integral domain, a characterization of trivial extension rings over which every ideal is homogeneous is given (see [9, Theorem 3.3 and Corollary 3.4]). Our aim in the remainder of this section is to extend this study to ntrivial extensions. It is worth noting that in the classical case (where n = 1), ideals J with Π0 (J) = 0 are homogeneous. This shows that the condition that all ideals J with Π0 (J) 6= 0 are homogeneous implies that all ideals of R ⋉1 M are homogeneous. In the context of R ⋉n M for n ≥ 2 we show that more situations can occur. Let us begin with the class of ideals J of R ⋉n M with Π0 (J) ∩ S 6= ∅ for a given subset S of regular elements of R.

Recall that a ring S is said to be présimplifiable if, for every a and b in S: ab = a implies a = 0 or b ∈ U (S). Présimplifiable rings were introduced and studied by Bouvier in a series of papers (see references) and they have also been investigated in [7, 8]. In [9], the notion of a présimplifiable ring is used when homogeneous ideals of the classical trivial extensions were studied. For example, we have that if R is présimplifiable but not an integral domain, then every ideal of R ⋉1 M is homogeneous if and only if M1 = 0 (see [9, Theorems 3.3 (4)]). This is why we first consider just subsets of regular elements. Theorem 5.4 Let S be a nonempty subset of R − Z(R) and let I be the class of ideals J of R ⋉n M with Π0 (J) ∩ S 6= ∅. Then the following assertions are equivalent. 1. Every ideal in I is homogeneous. 2. Every principal ideal in I is homogeneous. 3. For every s ∈ S and i ∈ {1, ..., n}, sMi = Mi . 4. Every principal ideal h(s, m1 , ..., mn )i with s ∈ S has the form I ⋉n M where I is a principal ideal of R with I ∩ S 6= ∅. 5. Every ideal in I has the form I ⋉n M where I is an ideal of R with I ∩ S 6= ∅. Proof. (1) ⇒ (2). Obvious. (2) ⇒ (3). Let s ∈ S and i ∈ {1, ..., n}. We only need to prove that Mi ⊆ sMi . Consider an element mi of Mi . Since s ∈ S, h(s, 0, ..., 0, mi , 0, ..., 0)i is homogeneous. Then (s, 0, ..., 0) ∈ h(s, 0, ..., 0, mi , 0, ..., 0)i, so there is (x, e1 , ..., en ) ∈ R ⋉n M1 ⋉ · · · ⋉ Mn such that (s, 0, ..., 0, mi , 0, ..., 0)(x, e1 , ..., en ) = (s, 0, ..., 0).


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Since s is regular, x = 1. Then mi = (−s)ei , as desired. (3) ⇒ (4). Let h(s, m1 , ..., mn )i be a principal ideal of R ⋉n M with s ∈ S. By (3), (s, m1 , ..., mn )(0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn ) = 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn . This implies that 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn ⊂ h(s, m1 , ..., mn )i. Using this inclusion and (3), we get 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn−1 ⋉ 0 ⊂ h(s, m1 , ..., mn )i. Then inductively we get 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mi ⋉ 0 ⋉ · · · ⋉ 0 ⊂ h(s, m1 , ..., mn )i for every i ∈ {1, ..., n}. Thus 0 ⋉n M1 ⋉ · · · ⋉ Mn ⊂ h(s, m1 , ..., mn )i. Therefore by Lemma 4.6 and Proposition 5.2 (2), h(s, m1 , ..., mn )i has the form I ⋉n M where I = sR. (4) ⇒ (5). Consider an ideal J in I . Then there is an element (s, m1 , ..., mn ) ∈ J such that s ∈ Π0 (J) ∩ S. Therefore using (4) and Lemma 4.6, we get the desired result. (5) ⇒ (1). Obvious. As an example, we can consider the trivial extension S := Z ⋉2 ZW ⋉ Q where ZW is the ring of fractions of Z with respect to the multiplicatively closed subset W = {2k |k ∈ N} of Z. Then the principal ideal h(3, 1, 0)i of S is not homogeneous. Deny, we must have (3, 0, 0) ∈ h(3, 1, 0)i. Thus there is (a, e, f ) ∈ S such that (3, 0, 0) = (3, 1, 0)(a, e, f ). But this implies that a = 1 and then e = −1 3 , which is absurd. The following result is an extension of [9, Theorem 3.9]. Recall that an ideal is said to be regular if it contains a regular element. Here, from Proposition 4.9, an ideal of R ⋉n M is regular if and only if it contains an element (s, m1 , ..., mn ) with s ∈ R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )). Corollary 5.5 Let S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )). Then the following assertions are equivalent. 1. Every regular ideal of R ⋉n M is homogeneous. 2. Every principal regular ideal of R ⋉n M is homogeneous. 3. For every s ∈ S and i ∈ {1, ..., n}, sMi = Mi (or equivalently, MiS = Mi ). 4. Every principal ideal h(s, m1 , ..., mn )i with s ∈ S has the form I ⋉n M where I is a principal ideal of R with I ∩ S 6= ∅. 5. Every regular ideal of R⋉n M has the form I ⋉n M where I is an ideal of R with I ∩S 6= ∅. Consequently, if R ⋉n M is root closed (in particular, integrally closed), then every regular ideal of R ⋉n M has the form given in (5). Proof. The proof is similar to the one of [9, Theorem 3.9].

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Compare the following result with [9, Corollary 3.4]. Corollary 5.6 Assume that R is an integral domain. Then the following assertions are equivalent. 1. Every ideal J of R ⋉n M with Π0 (J) 6= 0 is homogeneous. 2. Every principal ideal J of R ⋉n M with Π0 (J) 6= 0 is homogeneous. 3. For every i ∈ {1, ..., n}, Mi is divisible. 4. Every principal ideal h(s, m1 , ..., mn )i of R ⋉n M with s 6= 0 has the form I ⋉n M where I is a nonzero principal ideal of R. 5. Every ideal J of R ⋉n M with Π0 (J) 6= 0 has the form I ⋉n M where I is a nonzero ideal of R. 6. Every ideal of R ⋉n M is comparable to 0 ⋉n M . Proof. The equivalence (5) ⇔ (6) is a simple consequence of Lemma 4.6. The proof of Theorem 5.4 shows that another situation can be considered. This is given in the following result. We use AnnR (H) to denote the annihilator of an R-module H. Theorem 5.7 Let I be the class of ideals J of R ⋉n M with Π0 (J) ∩ S 6= ∅ where S is a nonempty subset of R − {0} such that, for every s ∈ S, AnnR (s) ⊆ AnnR (Mi ). Then the following assertions are equivalent. 1. Every ideal in I is homogeneous. 2. Every principal ideal in I is homogeneous. 3. For every s ∈ S and i ∈ {1, ..., n}, sMi = Mi . 4. Every principal ideal h(s, m1 , ..., mn )i with s ∈ S has the form I ⋉n M where I is a principal ideal of R with I ∩ S 6= ∅. 5. Every ideal in I has the form I ⋉n M where I is an ideal of R with I ∩ S 6= ∅. Proof. We only need to prove the implication (2) ⇒ (3). Let s ∈ S and i ∈ {1, ..., n} and consider an element mi of Mi − {0}. Since s ∈ S, h(s, 0, ..., 0, mi , 0..., 0)i is homogeneous. Then (s, 0, ..., 0) ∈ h(s, 0, ..., 0, mi , 0..., 0)i, so there is (x, e1 , ..., en ) ∈ R ⋉n M1 ⋉ · · · ⋉ Mn such that (s, 0, ..., 0, mi , 0..., 0)(x, e1 , ..., en ) = (s, 0, ..., 0). Then sx = s and, by the hypothesis on S, (x − 1)mi = 0. Therefore mi = xmi = (−s)ei , as desired.


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For an example of a ring that satisfies the condition of the previous result, consider a ring R with an idempotent e ∈ R − {1, 0} and set S = {e} and Mi = Re for every i ∈ {1, ..., n}. Thus, since eMi = Mi for every i ∈ {1, ..., n}, every ideal J of R ⋉n M with e ∈ Π0 (J) is homogeneous. Unlike the classical case (where n = 1), the fact that, for every i ∈ {1, ..., n}, Mi is divisible does not necessarily imply that every ideal is homogeneous. For that, we consider the 2-trivial extension S := k ⋉2 (k × k) ⋉ (k × k) where k is a field. Then the principal ideal h(0, (1, 0), (0, 1))i of S is not homogeneous. Indeed, if it were homogeneous, we must have (0, (1, 0), (0, 0)) ∈ h(0, (1, 0), (0, 1))i. Thus there is (a, (e, f ), (e′ , f ′ )) ∈ S such that (0, (1, 0), (0, 0)) = (0, (1, 0), (0, 1))(a, (e, f ), (e′ , f ′ )). But this implies that (a, 0) = (1, 0) and (0, 0) = (e, a), which is absurd. This example naturally leads us to investigate when every ideal J of R ⋉n M with Π0 (J) = 0 is homogeneous. In this context, the notion of a présimplifiable module is used. For that, recall that an R-module H is called R-présimplifiable if, for every r ∈ R and h ∈ H, rh = h implies h = 0 or r ∈ U (R). For example, over an integral domain, every torsion-free module is présimplifiable (see [7] and also [3]). In studying the question when every ideal J of R ⋉n M with Π0 (J) = 0 is homogeneous, several different cases occur. For this we use the following lemma. Lemma 5.8 Let J be an ideal of R⋉n M such that, for i ∈ {1, ..., n}, Π0 (J) = 0,...,Πi−1 (J) = 0, Πi (J) 6= 0. Then the following assertions are true. 1. For i = n, the ideal J is homogeneous and it has the form 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Πn (J). 2. For i 6= n, if 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mi+1 ⋉ · · · ⋉ Mn ⊂ J, then J is homogeneous and it has the form 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Πi (J) ⋉ Mi+1 ⋉ · · · ⋉ Mn . Proof. Straightforward. Theorem 5.9 Assume that n ≥ 2 and Mj is présimplifiable for a given j ∈ {1, ..., n − 1}. Let I be the class of ideals J of R ⋉n M with Πi (J) = 0 for every i ∈ {0, ..., j − 1} and Πj (J) 6= 0. Then the following assertions are equivalent. 1. Every ideal in I is homogeneous. 2. Every principal ideal in I is homogeneous. 3. For every k ∈ {j + 1, ..., n} and every mj ∈ Mj − {0}, Mk = mj Mk−j . 4. Every principal ideal h(0, 0, ..., 0, mj , ..., mn )i with mj 6= 0 has the form 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ N ⋉ Mj+1 ⋉ · · · ⋉ Mn where N is a nonzero cyclic submodule of Mj .

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D. D. Anderson et al. 5. Every ideal in I has the form 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ N ⋉ Mj+1 ⋉ · · · ⋉ Mn where N is a nonzero submodule of Mj . 6. Every ideal in I contains 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mj+1 ⋉ · · · ⋉ Mn .

Proof. The implication (3) =⇒ (4) is proved similarly to the implication (3) ⇒ (4) of Theorem 5.4. The implication (6) =⇒ (1) is a simple consequence of Lemma 5.8. Then only the implication (2) =⇒ (3) needs a proof. Let k ∈ {j + 1, ..., n}, mj ∈ Mj − {0} and mk ∈ Mk − {0}. Then the principal ideal p = h(0, ..., 0, mj , 0, ..., 0, mk , 0, ..., 0)i is homogeneous. This implies that (0, ..., 0, mj , 0, ..., 0) ∈ p and so there exists (r, e1 , ..., en ) ∈ R ⋉n M such that (0, ..., 0, mj , 0, ..., 0) = (r, e1 , ..., en )(0, ..., 0, mj , 0, ..., 0, mk , 0, ..., 0). Then rmj = mj and rmk + ek−j mj = 0. Since Mj is présimplifiable, r is invertible and then mk = −r −1 ek−j mj , as desired. For examples of rings that satisfy the conditions of the previous result, we can consider the following two 2-trivial extensions: Z ⋉2 ZW ⋉ Q and Z ⋉2 ZW ⋉ ZW where ZW is the ring of fractions of Z with respect to the multiplicatively closed subset W = {2k |k ∈ N} of Z. The following particular cases are of interest. Corollary 5.10 Assume that n ≥ 2 and Mn−1 is présimplifiable. Let I be the class of ideals J of R ⋉n M with Πi (J) = 0 for every i ∈ {0, ..., n − 2}. Then the following assertions are equivalent. 1. Every ideal in I is homogeneous. 2. For every mn−1 ∈ Mn−1 − {0}, Mn = mn−1 M1 . 3. Every ideal in I is comparable to 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn . Proof. (1) =⇒ (2). This is a particular case of the corresponding one in Theorem 5.9. (2) =⇒ (3). Let I be an ideal of R ⋉n M in I . If Πn−1 (I) 6= 0, then Theorem 5.9 shows that I contains 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn . Otherwise, Πn−1 (I) = 0 which means that 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ Mn contains I. (3) ⇒ (1). Let I be a nonzero ideal of R ⋉n M in I . If Πn−1 (I) 6= 0, then Theorem 5.9 shows that I is homogeneous. The other case is a consequence of the assertion (1) of Lemma 5.8.

When n = 2, we get the following particular case of Corollary 5.10. Corollary 5.11 Assume that M1 is présimplifiable and n = 2. Let I be the class of ideals J of R ⋉2 M with Π0 (J) = 0. Then the following assertions are equivalent.


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1. Every ideal in I is homogeneous. 2. For every m1 ∈ M1 − {0}, M2 = m1 M1 . 3. Every ideal in I is comparable to 0 ⋉2 0 ⋉ M2 . When j = 1 in Theorem 5.9, there are additional conditions equivalent to (1)-(6). The study of this case leads us to introduce the following notion in order to avoid trivial situations. Definition 5.12 Assume that n ≥ 2. For i ∈ {1, ..., n − 1} and j ∈ {2, ..., n} with ji ≤ n, Mi is said to be ϕ-j-integral (where ϕ = {ϕi,j } i+j≤n is the family of multiplications) if, for any j 1≤i,j≤n−1

elements mi1 , ..., mij of Mi , if the product mi1 · · · mij = 0, then at least one of the mik ’s is zero. If no ambiguity arises, Mi is simply called j-integral. Corollary 5.13 Assume that n ≥ 2, M1 is présimplifiable and k-integral for every k ∈ {2, ..., n− 1}. Let I be the class of ideals J of R ⋉n M with Π0 (J) = 0 and Π1 (J) 6= 0. Then the following assertions are equivalent. 1. Every ideal in I is homogeneous. 2. Every principal ideal in I is homogeneous. 3. For every k ∈ {2, ..., n} and every m1 ∈ M1 − {0}, Mk = m1 Mk−1 . 4. For every k ∈ {2, ..., n} and every nonzero elements m11 , ..., m1k−1 ∈ M1 − {0}, Mk = m11 · · · m1k−1 M1 . 5. For every k ∈ {2, ..., n} and every nonzero element m ∈ M1 − {0}, Mk = mk−1 M1 . 6. Every principal ideal h(0, m1 , ..., mn )i with m1 6= 0 has the form 0 ⋉n N ⋉ M2 · · · ⋉ Mn where N is a nonzero cyclic submodule of M1 . 7. Every ideal in I has the form 0 ⋉n N ⋉ M2 · · · ⋉ Mn where N is a nonzero submodule of M1 . 8. Every ideal in I contains 0 ⋉n 0 ⋉ M2 ⋉ · · · ⋉ Mn . Proof. The equivalences (3) ⇔ (4) ⇔ (5) are easily proved. The following result shows that, in fact, the conditions of Corollary 5.13 above are necessary and sufficient to show that every ideal J of R ⋉n M with Π0 (J) = 0 is homogeneous. Note that Corollary 5.11 presents the case n = 2. Thus in the following result we may assume that n ≥ 3. Corollary 5.14 Assume that n ≥ 3 and M1 is présimplifiable and k-integral for every k ∈ {2, ..., n − 1}. Then the following assertions are equivalent.

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D. D. Anderson et al. 1. Every ideal J of R ⋉n M with Π0 (J) = 0 and Π1 (J) 6= 0 is homogeneous. 2. For every j ∈ {1, ..., n − 1}, Mj is présimplifiable and every ideal J of R ⋉n M with Π0 (J) = 0 is homogeneous.

Proof. We only need to prove that (1) ⇒ (2). Let j ∈ {1, ..., n − 1} and consider mj ∈ Mj − {0}. Let r ∈ R such that rmj = mj . By Corollary 5.13 (4), there are m11 , ..., m1j ∈ M1 − {0} such that mj = m11 · · · m1j . Then rm11 · · · m1j = m11 · · · m1j which implies that (rm11 − m11 )m12 · · · m1j = 0. Now, since M1 is k-integral for every k ∈ {2, ..., n − 1}, rm11 − m11 = 0. Therefore r is invertible since M1 is présimplifiable. So Mj is présimplifiable. Now, to prove that every ideal J of R⋉n M with Π0 (J) = 0 is homogeneous, it suffices to prove that Mk = mj Mk−j for every k ∈ {2, ..., n}, every j ∈ {1, ..., k − 1} and every mj ∈ Mj − {0} (by Theorem 5.9). The case where k = 2 is trivial. Thus fix k ∈ {3, ..., n} and j ∈ {1, ..., k − 1}. Consider mj ∈ Mj − {0} and mk ∈ Mk − {0}. We prove that mk = mj mk−j for some mk−j ∈ Mk−j − {0}. By Corollary 5.13 (5), mj = mj−1 m1 for some m, m1 ∈ M1 − {0}. And, by Corollary 5.13 (3), mk = m1 mk−1 for some mk−1 ∈ Mk−1 − {0}. Also, by Corollary 5.13 (5), mk−1 = mk−2 m′1 for some m′1 ∈ M1 − {0}. Then mk = mk−2 m1 m′1 = (mj−1 m1 )(mk−j−1 m′1 ) = mj mk−j where mk−j = mk−j−1 m′1 ∈ Mk−j − {0}, as desired. Finally, we give a case when we can characterize rings in which every ideal is homogeneous. Note that, when R is a ring with aMi = Mi for every i ∈ {1, ..., n − 1} and every a ∈ R − {0}, and Mi = mi−1 M1 for every i ∈ {2, ..., n} and every nonzero element m ∈ M1 − {0}, then R is an integral domain and Mi must be torsion-free for every i ∈ {1, ..., n − 1}. Corollary 5.15 Suppose that n ≥ 2 and R is an integral domain. Assume that Mi is torsionfree, for every i ∈ {1, ..., n − 1}, and that M1 is k-integral for every k ∈ {2, ..., n − 1}. Then the following assertions are equivalent. 1. Every ideal of R ⋉n M is homogeneous. 2. The following two conditions are satisfied: i. For every i ∈ {1, ..., n}, Mi is divisible, and ii. For every i ∈ {2, ..., n} and every m1 ∈ M1 − {0}, Mi = m1 Mi−1 . Proof. Simply use Corollaries 5.6 and 5.13 and Theorem 5.9. It is easy to show that the two n-trivial extensions Z ⋉n Q ⋉ · · · ⋉ Q and Z ⋉n Q ⋉ · · · ⋉ Q ⋉ Q/Z satisfy the conditions of the above result and so every ideal of these rings is homogeneous. We end this section with the following particular case.


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Corollary 5.16 Suppose that n ≥ 2. Consider the n-trivial extension S := k ⋉n E1 ⋉ · · · ⋉ En where k is a field and, for i ∈ {1, ..., n}, Ei is a k-vector space. Suppose that E1 is k-integral for every k ∈ {2, ..., n − 1}. Then the following assertions are equivalent. 1. Every ideal of S is homogeneous. 2. For every k ∈ {2, ..., n}, every j ∈ {1, ..., k − 1} and every ej ∈ Ej − {0}, Ek = ej Ek−j . As a particular case, we can consider a field extension K ⊆ F , then every ideal of S := K ⋉n F ⋉ · · · ⋉ F is homogeneous. Namely, every proper ideal of S has the form 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ N ⋉ F ⋉ · · · ⋉ F where N is a K-subspace of F .

6

Some ring-theoritic properties of R ⋉n M

In this section, we determine when R ⋉n M has certain ring properties such as being Noetherian, Artinian, Manis valuation, Pr¨ ufer, chained, arithmetical, a π-ring, a generalized ZPI-ring or a PIR. We end the section with a remark on a question posed in [2] concerning m-Boolean rings. We begin by characterizing when the n-trivial extensions are Noetherian (resp., Artinian). The following result extends [9, Theorem 4.8]. Theorem 6.1 The ring R ⋉n M is Noetherian (resp., Artinian) if and only if R is Noetherian (resp., Artinian) and, for every i ∈ {1, ..., n}, Mi is finitely generated. Proof. Similar to the proof of [9, Theorem 4.8]. The following result is an extension of [9, Theorem 4.2 and Corollary 4.3]. It investigates the integral closure of R ⋉n M in the total quotient ring T (R ⋉n M ) of R ⋉n M . Theorem 6.2 Let S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )). If R′ is the integral closure of R in T (R), then (R′ ∩ RS ) ⋉n M1S ⋉ · · · ⋉ MnS is the integral closure of R ⋉n M in T (R ⋉n M ). In particular, 1. If R is an integrally closed ring, then R ⋉n M1S ⋉ · · · ⋉ MnS is the integral closure of R ⋉n M1 ⋉ · · · ⋉ Mn in T (R ⋉n M1 ⋉ · · · ⋉ Mn ), and 2. If Z(Mi ) ⊆ Z(R) for all i ∈ {1, ..., n}, then R ⋉n M1S ⋉ · · · ⋉ MnS is integrally closed if and only if R is integrally closed. Proof. All statements can be proved similarly to the corresponding ones of [9, Theorem 4.2 and Corollary 4.3].

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It is worth noting as in the classical case that R ⋉n M can be integrally closed without R being integrally closed (see the example given after [9, Corollary 4.3]). Similar to the classical case [9, Theorem 4.16 (1) and (2)], as a consequence of Theorem 6.2 and Corollary 5.5, we give the following result which characterizes when R ⋉n M is (Manis) valuation and when it is Pr¨ ufer. First, recall these two notions. Let S be a subring of a ring T , and let P be a prime ideal of S. Then (S, P ) is called a valuation pair on T (or just S is a valuation ring on T ) if there is a surjective valuation v : T −→ G ∪ {∞} (v(xy) = v(x) + v(y), v(x + y) ≥ min{v(x), v(y)}, v(1) = 0 and v(0) = ∞) where G is a totally ordered abelian group, with S = {x ∈ T |v(x) ≥ 0} and P = {x ∈ T |v(x) > 0}. This is equivalent to if x ∈ T − S, then there exists x′ ∈ P with xx′ ∈ S − P . A valuation ring S is called a (Manis) valuation ring if T = T (S). Also, S is called a Pr¨ ufer ring if every finitely generated regular ideal of S is invertible. This is equivalent to every overring of S being integrally closed (see [38] for more details). Corollary 6.3 Let S = R − (Z(R) ∪ Z(M1 ) ∪ · · · ∪ Z(Mn )). 1. R⋉n M is a Manis valuation ring if and only if R is a valuation ring on RS and Mi = MiS for every i ∈ {1, ..., n}. 2. R ⋉n M is a Pr¨ ufer ring if and only if, for every finitely generated ideal I of R with I ∩ S 6= ∅, I is invertible and Mi = MiS for every i ∈ {1, ..., n}. Now, as an extension of [9, Theorem 4.16 (3)], we characterize when R ⋉n M is a chained ring. Recall that a ring S is said to be chained if the set of ideals of S is totally ordered by inclusion. As an exception to Convention 4.2, in the following results (Lemma 6.4, Theorem 6.5, Corollary 6.6, Lemma 6.7 and Theorem 6.7), a module in the family associated to an n-trivial extension can be zero. The proof of the desired result uses the following lemma which gives another characterization of a particular n-trivial extension with the property that every ideal is homogeneous. Lemma 6.4 Assume that R is quasi-local with maximal ideal m. Suppose also that at least one of the modules of the family M is nonzero. Then every ideal of R ⋉n M is homogeneous if and only if the following three conditions are satisfied: 1. R is an integral domain. 2. For every i ∈ {1, ..., n}, Mi is divisible. 3. For every 1 ≤ i ≤ j ≤ n (when n ≥ 2), if Mi 6= 0 and Mj 6= 0, then Mj−i 6= 0 and eMi = Mj for every e ∈ Mj−i . In this case, each ideal has one of the forms I ⋉n M , for some ideal I of R, or 0 ⋉n 0 ⋉ · · · ⋉ 0 ⋉ N ⋉ Mj+1 ⋉ · · · ⋉ Mn where N is a nonzero submodule of Mj for some j ∈ {1, ..., n}.


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Proof. =⇒ Clearly the first assertion is a simple consequence of the second one. Then we only need to prove the second and the third assertions. (2). Let r ∈ R − {0} and i ∈ {1, ..., n}. Consider an element mi ∈ Mi . If r 6∈ m, then r is invertible and trivially we get the result. Then assume r ∈ m. By hypothesis, the ideal h(r, 0, ..., 0, mi , 0, ..., 0)i is homogeneous, so there is (r ′ , m′1 , ..., m′n ) such that (r, 0, ..., 0) = (r, 0, ..., 0, mi , 0, ..., 0)(r ′ , m′1 , ..., m′n ). Then rr ′ = r and 0 = rm′i + r ′ mi . Thus r ′ cannot be in m, so r ′ is invertible and thus mi = −(r ′ )−1 rm′i , as desired. (3). Let 1 ≤ i ≤ j ≤ n such that Mi 6= 0 and Mj 6= 0. Consider mi ∈ Mi − {0} and mj ∈ Mj − {0}. By hypothesis, h(0, ..., 0, mi , 0, ..., 0, mj , 0, ..., 0)i is homogeneous. Then (0, ..., 0, mj , 0, ..., 0) = (0, ..., 0, mi , 0, ..., 0, mj , 0, ..., 0)(r ′ , m′1 , ..., m′n ) for some (r ′ , m′1 , ..., m′n ) ∈ R ⋉n M . This implies that r ′ mi = 0 and

r ′ mj + mi m′j−i = mj .

If Mj−i = 0, we get r ′ mi = 0 and (r ′ − 1)mj = 0. This is impossible since either r ′ or r ′ − 1 is invertible. Then Mj−i 6= 0. Now, suppose that r ′ 6= 0. By (2), there exists m′′j−i ∈ Mj−i such that m′j−i = r ′ m′′j−i . Hence using the fact that r ′ mi = 0, the equality r ′ mj + mi m′j−i = mj becomes r ′ mj = mj . As in the previous case, this is impossible. Therefore r ′ = 0 and this gives the desired result. ⇐= We only need to prove that every principal ideal h(s, m1 , ..., mn )i of R ⋉n M is homogeneous. For this, distinguish two cases s 6= 0 and s = 0 and follow an argument similar to that of (3) ⇒ (4) of Theorem 5.4. Theorem 6.5 Assume that n ≥ 2 and that at least one of the modules of the family M is nonzero. Then the ring R ⋉n M is chained if and only if the following conditions are satisfied: 1. R is a valuation domain, 2. For every i ∈ {1, ..., n}, Mi is divisible, 3. For every 1 ≤ i ≤ j ≤ n, if Mi 6= 0 and Mj 6= 0, then Mj−i 6= 0 and eMi = Mj for every e ∈ Mj−i , and 4. For every i ∈ {1, ..., n}, the set of all (cyclic) submodules of Mi is totally ordered by inclusion.

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Proof. =⇒ First, we prove that R is a chained ring. Consider two ideals I and J of R. Then I ⋉n M and J ⋉n M are two ideals of R ⋉n M . Then they are comparable and so are I and J as desired. A similar argument can be used to prove the last assertion. Now, we prove that every ideal of R ⋉n M is homogeneous. Then by Lemma 6.4, we get the other assertions. Consider a nonzero ideal K of R ⋉n M . If Π0 (K) 6= 0, then necessarily 0 ⋉n M ⊂ K. Then by Lemma 4.6, K is homogeneous. Now, let i ≥ 1 be the smallest integer such that Πi (K) 6= 0. If i = n, then by the first assertion of Lemma 5.8, K is homogeneous. If i 6= n, then necessarily 0 ⋉n · · · ⋉ 0 ⋉ Mi+1 ⋉ · · · ⋉ Mn ⊂ K. Thus by the second assertion of Lemma 5.8, K is homogeneous, as desired. ⇐= Using Lemma 5.8, we deduce that any two ideals I and J of R ⋉n M have the forms I = 0 ⋉n · · · ⋉ 0 ⋉ Ii ⋉ Mi+1 ⋉ · · · ⋉ Mn and J = 0 ⋉n · · · ⋉ 0 ⋉ Jj ⋉ Mj+1 ⋉ · · · ⋉ Mn for some i, j ∈ {0, ..., n} where Ii and Jj are submodules of Mi and Mj respectively (here M0 = R). If i 6= j, then obviously I and J are comparable. If i = j, then using the first and the last assertion, we can show that Ii and Jj are comparable and so are I and J, as desired. Using Theorem 6.5 and Corollary 3.8, we get an extension of [9, Theorem 4.16 (4)] which characterizes when R ⋉n M is arithmetical. Recall that a ring S is arithmetical if and only if SP is chained for each prime (maximal) ideal P of S. Also, recall that, for a ring S, an S-module H is called arithmetical if, for each prime (maximal) ideal P of S, the set of submodules of HP is totally ordered by inclusion. Finally, recall the support of an S-module H, supp(H), over a ring S is the set of all prime ideals P of S such that HP 6= 0. Corollary 6.6 The ring R ⋉n M is arithmetical if and only if the following conditions are satisfied: 1. R is arithmetical, 2. For every i ∈ {1, ..., n}, Mi is an arithmetical R-module, 3. For every P ∈ ∪supp(Mi ), RP is a valuation domain, i

4. For every i ∈ {1, ..., n} and every P ∈ supp(Mi ), MiP is a divisible RP -module, and 5. For every 1 ≤ i ≤ j ≤ n, if P ∈ supp(Mi ) ∩ supp(Mj ), then P ∈ supp(Mj−i ) and eMiP = MjP for every e ∈ M(j−i)P . Recall that a ring S is called a generalized ZPI-ring (resp., a π-ring) if every proper ideal (resp., proper principal ideal) of S is a product of prime ideals. An integral domain which is a π-ring is called a π-domain. Clearly, a generalized ZPI-domain is nothing but a Dedekind domain. It is well known (for example, see [28, Sections 39 and 46]) that S is a π-ring (resp., a generalized ZPI-ring, a principal ideal ring (PIR)) if and only if S is a finite direct product


39

of the following types of rings: (1) π-domains (resp., Dedekind domains, PIDs) which are not fields, (2) special principal ideal rings (SPIRs) and (3) fields. Our next results extend [9, Lemma 4.9 and Theorem 4.10]. They characterize when R ⋉n M is a π-ring, a generalized ZPI-ring or a PIR. Lemma 6.7 If R ⋉n M is a π-ring (resp., a generalized ZPI-ring, a PIR), then R is a π-ring (resp., a generalized ZPI-ring, a PIR). Hence R = R1 × · · · × Rs where Ri is either (1) a πdomain (resp., a Dedekind domain, a PID) but not a field, (2) an SPIR, or (3) a field. Let Mj,i = (0 × · · · 0 × Rj × 0 × · · · × 0)Mi where 1 ≤ i ≤ n and 1 ≤ j ≤ s. If Ri is a domain or SPIR, but not a field, then Mj,i = 0 while if Ri is a field, Mj,i = 0 or Mj,i ∼ = Ri . Conversely, if R = R1 × · · · × Rs and Mi = M1,i × · · · × Ms,i are as above and R is a π-ring (resp., a generalized ZPI-ring, a PIR), then R ⋉n M is a π-ring (resp., a generalized ZPI-ring, a PIR). Proof. Using Theorem 3.9, the proof is similar to that of [9, Lemma 4.9]. Theorem 6.8 R ⋉n M is a π-ring (resp., a generalized ZPI-ring, a PIR) if and only if R is a π-ring (resp., a generalized ZPI-ring, a PIR) and Mi is cyclic with annihilator Pi1 · · · Pis where Pi1 , ...,Pis are some idempotent maximal ideals of R (if is = 0, Ann(Mi ) = R, that is, Mi = 0). Proof. Similar to the proof of [9, Theorem 4.10]. We end the section with a remark on a question posed in [2]. Recall that a ring R is called m-Boolean for some m ∈ N, if char R = 2 and x1 x2 · · · xm (1 + x1 ) · · · (1 + xm ) = 0 for all x1 , ..., xm ∈ R. Thus Boolean rings are just 1-Boolean rings. It is shown in [2, Theorem 10] that 2-Boolean rings can be represented as trivial extensions. Namely, it is proved that if R is 2-Boolean, then R ∼ = B ⋉ Nil(R) where B = {b ∈ R|b2 = b} ([2, Theorem 10]). Based on this result the following natural question is posed (see [2, page 74]): Whether [2, Theorem 10] can be extended to m-Boolean rings for m ≥ 2? One can ask whether the n-trivial extension is the suitable construction to solve this question. Using [2, Theorem 6], one can show that the amalgamated algebras along an ideal (introduced in [34]) resolve partially this question. Recall that, given a ring homomorphism f : A −→ B and an ideal J of B, the amalgamation of A with B along J with respect to f is the following subring of A × B: A ⊲⊳f B = {(a, f (a) + j)|a ∈ A, j ∈ J}. ˙ where A⊕J ˙ ⊆ A× B is the ring whose underlying group is A⊕ J with Note that A ⊲⊳f B ∼ = A⊕J ′ ′ multiplication given by (a, x)(a , x ) = (aa′ , ax′ + a′ x + xx′ ) for all a, a′ ∈ A and x, x′ ∈ J. Here J is an A-module via f and then ax′ := f (a)x′ and a′ x := f (a′ )x (see [34] for more details). Now, if R is m-Boolean for m ≥ 2, then from [2, Theorems 6 and 7], R = B ⊕ Nil(R) where B = {b ∈ R|b2 = b}. Then ∼ ˙ R∼ = B ⊕Nil(R) = B ⊲⊳i Nil(R)

40


where i : B ֒→ R is the canonical injection. Actually any n-trivial extension R ⋉n M can be seen as the amalgamation of R with R ⋉n M along 0 ⋉n M with respect to the canonical injection. This leads to pose the following question for every m ≥ 2: Is any m-Boolean ring an m-trivial extension?

7

Divisibility properties of R ⋉n M

Factorization in commutative rings with zero divisors was first investigated in a series of papers by Bouvier, Fletcher and Billis (see References), where the focus had been on the unicity property. The papers [1, 7, 8] marked the start of a systematic study of factorization in commutative rings with zero divisors. Since then, this theory has attracted the interest of a number of authors. The study of divisibility properties of the classical trivial extension has lead to some interesting examples and then to answering several questions (see [9, Section 5]). In this section we are interested in extending a part of this study to the context of n-trivial extensions. First, we recall the following definitions. Let S be a commutative ring and H an S-module. Two elements e, f ∈ H are said to be associates (written e ∼ f ) (resp., strong associates (written e ≈ f ), very strong associates (written e ∼ = f )) if Se = Sf (resp., e = uf for some u ∈ U (S), e ∼ f and either e = f = 0 or e = rf implies r ∈ U (S)). Taking H = S gives the notions of “associates” in S. We say that H is strongly associate if for every e, f ∈ H, e ∼ f ⇒ e ≈ f . When S is strongly associate as an S-module, we also say that S is strongly associate. Finally, recall that H is said to be S-présimplifiable if for r ∈ S and e ∈ H, re = e ⇒ r ∈ U (S) or e = 0. If S is S-présimplifiable we only say that S is présimplifiable. We begin with an extension of [9, Theorem 5.1]. Proposition 7.1 Let R ⊆ S be a ring extension such that U (S) ∩ R = U (R). 1. If S is présimplifiable, then every R-submodule of S is présimplifiable. In particular, R is présimplifiable. 2. Suppose that S = R ⊕ N as an R-module where N is a nilpotent ideal of S which satisfies either N 2 = 0 or N = ⊕ Ni as an R-module where S = R ⊕ N1 ⊕ N2 ⊕ · · · is a graded i∈N

ring. Then S is présimplifiable if R is présimplifiable and N is R-présimplifiable.

Proof. 1. Let H be an R-submodule of S. Consider e = xe with e ∈ H − {0} and x ∈ R − {0}. Since S is présimplifiable, x ∈ U (S) and so x ∈ U (S) ∩ R = U (R). 2. Let x = rx + nx 6= 0 and y = ry + ny be two elements of R ⊕ N = S where rx , ry ∈ R and nx , ny ∈ N , such that x = yx. Assume that rx 6= 0. Then rx = ry rx implies that ry ∈ U (R) ⊆ U (S), and, since N is nilpotent, y = ry + ny is invertible in S, as desired. Next, assume now that rx = 0. Then nx 6= 0 and so nx = ry nx + ny nx . In the case N 2 = 0, we have


41

nx = ry nx . Hence ry ∈ U (R) since N is présimplifiable, and as above y ∈ U (S). Finally, in the case where S = R ⊕ N1 ⊕ N2 ⊕ · · · is a graded ring, we may set nx = ni1 + · · · + nim with {i1 , ..., im } ⊂ N and m ∈ N such that i1 ≤ · · · ≤ im and ni1 6= 0. Then ni1 = ry ni1 which implies that ry ∈ U (R) and similarly as above y ∈ U (S). Proposition 7.2 Let R = ⊕ Ri be a graded ring. i∈N0

1. If R is strongly associate, then R0 is a strongly associate ring and Ri is a strongly associate R0 -module for every i ∈ N. 2. Suppose there exists n ∈ N such that Ri = 0 for every i ≥ n + 1, that is, R = R0 ⋉n R1 ⋉ · · · ⋉ Rn , and assume that R0 is a présimplifiable ring and R1 , ..., Rn−1 are présimplifiable R0 -modules. Then R is strongly associate if and only if Rn is strongly associate. Proof. 1. Let xi , yi ∈ Ri − {0} for i ∈ N0 such that R0 xi = R0 yi . Then Rxi = Ryi . Hence there is u = u0 + u1 + · · · ∈ U (R) such that xi = uyi . Then u0 ∈ U (R0 ) and xi = u0 yi , as desired. 2. Let x = xm +· · ·+xn and y = ym +· · ·+yn be two associate elements of R where m ∈ {0, ..., n} and xi , yi ∈ Ri for i ∈ {m, ..., n} such that xm and ym are nonzero. Then xm ∼ ym . In particular, there is α = α0 + · · · + αn such that x = αy. Then xm = α0 ym . Hence two cases occur. Case m 6= n. Since Rm is présimplifiable, α0 ∈ U (R0 ). Then α ∈ U (R), as desired. Case m = n (i.e., x = xm and y = ym ). Here, the result follows since Rn is strongly associate. Now we can give the extension of [9, Theorem 5.1] to the context of n-trivial extensions. Corollary 7.3 The following assertions are true. 1. R ⋉n M1 ⋉ · · · ⋉ Mn is présimplifiable if and only if R, M1 , ..., Mn are présimplifiable. 2. If R ⋉n M1 ⋉ · · · ⋉ Mn is strongly associate, then R, M1 , ..., Mn are strongly associate. 3. Suppose that R, M1 , ..., Mn are présimplifiable. Then R ⋉n M1 ⋉ · · · ⋉ Mn is strongly associate if and only if Mn is strongly associate. Now we investigate the extension of [9, Theorem 5.4]. It is convenient to recall the following definitions. Let S be a commutative ring. A nonunit a ∈ S is said to be irreducible or an atom (resp., strongly irreducible, very strongly irreducible) if a = bc implies a ∼ b or a ∼ c (resp., a ≈ b or a ≈ c, a ∼ = b or a ∼ = c) and a is said to be m-irreducible if Sa is a maximal element of the set of proper principal ideals of S. Note that, for a nonzero nonunit a ∈ S, a very strongly irreducible ⇒ a is m-irreducible ⇒ a is strongly irreducible ⇒ a is irreducible, but none of these implications can be reversed. In the case of an S-module H, we say that e ∈ H is S-primitive (resp., strongly S-primitive, very strongly S-primitive) if for a ∈ S and f ∈ H, e = af ⇒ e ∼ f (resp., e ≈ f , e ∼ = f ). And e is S-superprimitive if be = af for a, b ∈ S and

42


f ∈ H, implies a | b. Note that (1) e is S-primitive ⇔ Se is a maximal cyclic S-submodule of H, (2) e is S-superprimitive ⇒ e is very strongly S-primitive ⇒ e is strongly S-primitive ⇒ e is S-primitive, (3) if Ann(e) = 0, e is S-primitive ⇒ e is very strongly S-primitive, and (4) e is S-superprimitive ⇒ Ann(e) = 0. In the following results the homogeneous element (0, ..., 0, mi , 0, ..., 0) ∈ R ⋉n M where i ∈ {1, ..., n} and mi ∈ Mi − {0}, is denoted by mi . The following result extends [9, Theorem 5.4 (1)]. Proposition 7.4 Let i ∈ {1, ..., n} and mi , ni ∈ Mi − {0}. Then mi ∼ ni (resp., mi ≈ ni , mi ∼ = ni ) in Mi if and only if mi ∼ ni (resp., mi ≈ ni , mi ∼ = ni ) in R ⋉n M . Proof. The assertion is proved similarly to the corresponding classical one. It is worth noting that the analogue of the assertion (4) of [9, Theorem 5.4] does not hold in the context of n-trivial extensions with n ≥ 2. Indeed, consider the 2-trivial extension S = Z4 ⋉2 Z4 ⋉ Z4 . It is easy to show that ¯1 is superprimitive in the Z4 -module Z4 . However, (¯0, ¯0, ¯1) is not very strongly irreducible in S (since (¯2, ¯1, ¯2)2 = (¯0, ¯0, ¯1)). Moreover, even if we assume that R is an integral domain, we still don’t have the desired analogue. For this, take S = Z ⋉2 Z ⋉ Z. We have 1 is superprimitive in the Z-module Z. However, (0, 0, 1) is not very strongly irreducible in S (since (0, 1, 0)2 = (0, 0, 1)). The last example also shows that the assertion (2) of [9, Theorem 5.4] does not hold in the context of n-trivial extensions with n ≥ 2. Namely, if 0 6= mi = mj mk where (mi , mj , mk ) ∈ Mi × Mj × Mk , i ≥ 2 and j, k ∈ {1, ..., i − 1} with j + k = i, then mi cannot be irreducible. Indeed, mi = mj mk but neither mj nor mk are in hmi i ⊆ 0 ⋉n · · · ⋉ 0 ⋉ Mi ⋉ · · · ⋉ Mn . To extend [9, Theorem 5.1 (2)], we need to introduce the following definitions. Definition 7.5 Assume n ≥ 2 and each multiplication in the family ϕ = {ϕi,j }

i+j≤n

is

1≤i,j≤n−1

not trivial. Let i ∈ {2, ..., n}. An element mi ∈ Mi − {0} is said to be ϕ-indecomposable (or indecomposable relative to the family of multiplications ϕ) if, for every (mj , mk ) ∈ Mj × Mk (where j, k ∈ {1, ..., i − 1} with j + k = i), mi 6= mj mk . If no ambiguity can arise, ϕindecomposable elements are simply called indecomposables. For example, in Z ⋉2 Z ⋉ Q, every element in Q − Z is indecomposable. However, every element x ∈ Z (Z as a submodule of Q) is decomposable (since (0, 1, 0)(0, x, 0) = (0, 0, x)). Definition 7.6 Let i ∈ {2, ..., n}. The R-module Mi is said to be ϕ-integral (or integral relative to the family of multiplications ϕ) if, for every (mj , mk ) ∈ Mj × Mk (where j, k ∈ {1, ..., i − 1} with j + k = i), mj mk = 0 implies that mj = 0 or mk = 0. If no ambiguity can arise, a ϕ-integral R-module is simply called integral.


43

For example, for Z ⋉2 Z ⋉ Z, M2 = Z is integral. And, for Z ⋉2 Z ⋉ Z/2Z, Z/2Z is not integral since, for instance, ϕ1,1 (1, 2) = 1 2 = 0. Proposition 7.7 Assume n ≥ 2. Let i ∈ {1, ..., n} and mi ∈ Mi − {0}. If mi is irreducible (resp., strongly irreducible, very strongly irreducible) in R ⋉n M , then mi is primitive (resp., strongly primitive, very strongly primitive) in Mi . Conversely, three cases occur: Case i = 1. The reverse implication holds if R is an integral domain and Mj is torsion-free for every j ∈ {2, ..., n}. Case i = 2 (here n ≥ 2). The reverse implication holds if R is an integral domain, Mj is torsion-free for every j ∈ {1, ..., n} − {2} and m2 is indecomposable. Case i ≥ 3 (here n ≥ 3). The reverse implication holds if R is an integral domain, Mj is torsion-free for every j ∈ {1, ..., n} − {i}, Mj is integral for every j ∈ {2, ..., i − 1}, and mi is indecomposable. Proof. We only prove the primitive (irreducible) case. The two other cases are proved similarly. =⇒ Suppose that mi is irreducible and let mi = ani for some a ∈ R and ni ∈ Mi . Then mi = (a, 0, ..., 0)ni and so (R ⋉n M )mi = (R ⋉n M )ni . This implies that Rmi = Rni , as desired. ⇐= Let mi = (aj )(nj ) for some (aj ), (nj ) ∈ R ⋉n M . Then a0 n0 = 0. First, we show that the case a0 = 0 and n0 = 0 is impossible. Cases i = 1, 2 are easy and are left to the reader. So assume i ≥ 3. Suppose that a0 = 0 and n0 = 0. Then we have the following equalities: For j ∈ {2, ..., i− 1}, a1 nj−1 + a2 nj−2 + · · ·+ aj−1 n1 = 0 and a1 ni−1 + a2 ni−2 + · · ·+ ai−1 n1 = mi . A recursive argument on these equalities shows that, for l ∈ {2, ..., i}, there is k ∈ {0, ..., l − 1} such that (a0 , ..., ak ) = (0, ..., 0) and (n0 , ..., nl−(k+1) ) = (0, ..., 0). Indeed, it is clear this is true for l = 2. Then suppose this is true for a given l ∈ {2, ..., i−1}. So the equality a1 nl−1 +a2 nl−2 + · · · + al−1 n1 = 0 becomes ak+1 nl−(k+1) = 0. Then since Ml is integral, we get the desired result for l + 1. Thus for l = i, we get ak+1 ni−(k+1) = mi which is absurd since mi is indecomposable. Now, we may assume that a0 6= 0 and n0 = 0, so a0 n1 = 0. Since M1 is torsion-free, n1 = 0. Recursively we get nj = 0 for j ∈ {1, ..., i − 1}. Then a0 ni = mi , and since mi is primitive, there exists b0 ∈ R such that ni = b0 mi . It remains to show that there is bj ∈ Mj for j ∈ {1, ..., n − i} such that ni+j = bj mi and this implies that (ni ) = (0, ..., 0, ni , ni+1 , ...) = (b0 , ..., bn−i , 0, ..., 0)mi , as desired. We have a0 ni+1 + a1 ni = 0. Then using both a0 ni = mi and ni = b0 mi , we get a0 ni+1 + a1 b0 a0 ni = 0. Then a0 (ni+1 + a1 b0 ni ) = 0, so ni+1 + a1 b0 ni = 0 (since Mi+1 is torsionfree). Then ni+1 = −a1 b20 mi . Then we set b1 = −a1 b20 and so ni+1 = b1 mi . Similarly, using the equality a0 ni+2 + a1 ni+1 + a2 ni = 0 with the equalities a0 ni = mi and ni = b0 mi , we get a0 ni+2 + a1 b1 a0 ni + a2 b0 a0 ni = 0. So ni+2 + a1 b1 ni + a2 b0 ni = 0, then ni+2 = b2 mi where b2 = −a1 b1 b0 − a2 b20 . Finally, a recursive argument gives the desired result.

44


The following result extends [9, Theorem 5.4 (3)]. Proposition 7.8 Suppose that R has a nontrivial idempotent. Then for every i ∈ {1, ..., n} and mi ∈ Mi − {0}, mi is not irreducible in R ⋉n M . Proof. The assertion is proved similarly to the corresponding classical one. Now we are interested in some factorization properties. Recall that a ring S is called atomic if every (nonzero) nonunit of S is a product of irreducible elements (atoms) of S. Note that, as in the domain case, ACCP implies atomic. We begin with an extension of [9, Theorem 5.5 (2)] which characterizes when a trivial extension of a ring satisfies ACCP. For this, we need the following lemma. Lemma 7.9 Let i ∈ {0, ..., n} and consider two elements a = (0, ..., 0, ai , ai+1 , ..., an ) and b = (0, ..., 0, bi , bi+1 , ..., bn ) of R ⋉n M with ai 6= 0. Then the implication “hai ( hbi ⇒ bi 6= 0 and hai i ( hbi i” is true if either (1) 0 ≤ i ≤ n − 1 and Mi is présimplifiable (here M0 = R) or (2) i = n. Proof. Since hai ( hbi, there is c = (c0 , ..., cn ) ∈ R ⋉n M − U (R ⋉n M ) such that a = cb. Then ai = c0 bi and c0 ∈ / U (R). This shows that hai i ( hbi i in both cases. Theorem 7.10 Assume n ≥ 2. Suppose that Mi is présimplifiable for every i ∈ {0, ..., n − 1} (here M0 = R). Then R ⋉n M satisfies ACCP if and only if R satisfies ACCP and, for every i ∈ {1, ..., n}, Mi satisfies ACC on cyclic submodules. Proof. The proof of the direct implication is easy. Let us prove the converse. Suppose that R ⋉n M admits a strictly ascending chain of principal ideals h(a1,i )i ( h(a2,i )i ( · · · If there exists j0 ∈ N such that aj0 ,0 6= 0. Then for every k ≥ j0 , ak,0 6= 0. Then by Lemma 7.9, we get the following strictly ascending chain of principal ideals of R: haj0 ,0 i ( haj0 +1,0 i ( · · · This is absurd since R satisfies ACCP. Now, suppose that aj,0 = 0 for every j ∈ N and that there exists j1 ∈ N such that aj1 ,1 6= 0. Also, by Lemma 7.9, we obtain the following strictly ascending chain of cyclic submodules of M1 : haj1 ,1 i ( haj1 +1,1 i ( · · · This is absurd since M1 satisfies ACC on cyclic submodules. We continue in this way until the case where we may suppose that aj,i = 0 for every i ∈ {1, ..., n − 1} and every j ∈ N. Therefore, by Lemma 7.9, we get the desired result.


45

Now, we investigate when R ⋉n M is atomic. Namely, we give an extension of [9, Theorem 5.5 (4)]. Recall that an R-module N is said to satisfy MCC if every cyclic submodule of N is contained in a maximal (not necessarily proper) cyclic submodule of N . Theorem 7.11 Assume n ≥ 2. Suppose that Mi is présimplifiable for every i ∈ {0, ..., n − 1} (here M0 = R). Then R ⋉n M is atomic if R satisfies ACCP, Mi satisfies ACC on cyclic submodules, for every i ∈ {1, ..., n − 1}, and Mn satisfies MCC. Proof. The proof is slightly more technical than the one of [9, Theorem 5.5 (4)]. Here, we need to break the proof into the following n + 1 steps such that in the step number k ∈ {1, ..., n + 1} we prove that every nonunit element (mi ) ∈ R ⋉n M with m0 = 0,..., mk−2 = 0 and mk−1 6= 0 is a product of irreducibles. We use an inductive argument for the first n steps. Step 1. Suppose there is a nonunit element (mi ) of R ⋉n M with m0 6= 0 and such that (mi ) cannot be factored into irreducibles. Then there exist (a1,i ), (b1,i ) ∈ R ⋉n M − U (R ⋉n M ) such that (mi ) = (a1,i )(b1,i ) and neither (mi ) and (a1,i ) nor (mi ) and (b1,i ) are associate. Since 0 6= m0 = a1,0 b1,0 , a1,0 6= 0 and b1,0 6= 0. Clearly (a1,i ) or (b1,i ) must be reducible, say (a1,i ). Also, for (a1,i ) there are (a2,i ), (b2,i ) ∈ R ⋉n M − U (R ⋉n M ) such that (a1,i ) = (a2,i )(b2,i ) and neither (a1,i ) and (a2,i ) nor (a1,i ) and (b2,i ) are associate. As above, a2,0 6= 0 and b2,0 6= 0 and say (a2,i ) is reducible. So we continue and then we obtain a strictly ascending chain h(mi )i ( h(a1,i )i ( h(a2,i )i ( · · · Using Lemma 7.9, we get a strictly ascending chain of principal ideals of R hm0 i ( ha1,0 i ( ha2,0 i ( · · · This is absurd since R satisfies ACCP. Step j (1 ≤ j ≤ n). Suppose there is a nonunit element (mi ) ∈ R ⋉n M with m0 = 0,..., mj−2 = 0 and mj−1 6= 0 that is not a product of irreducibles. Then there are (a1,i ), (b1,i ) ∈ R ⋉n M − U (R ⋉n M ) such that (mi ) = (a1,i )(b1,i ) and neither (mi ) and (a1,i ) nor (mi ) and (b1,i ) are associate. Then a1,0 b1,j−1 + a1,1 b1,j−2 + · · · + a1,j−2 b1,1 + a1,j−1 b1,0 = mj−1 6= 0. If a1,k = 0 for every k ∈ {0, ..., j − 2}, then necessarily b1,0 6= 0. Hence by the preceding steps, (b1,i ) is a product of irreducibles and then by hypothesis on (mi ), (a1,i ) is reducible. If a1,k 6= 0 for some k ∈ {0, ..., j − 2}, then (a1,i ) is a product of irreducibles and (b1,i ) is reducible. Thus, by symmetry, we may assume that (a1,i ) is reducible and it is not a product of irreducibles. So, necessarily a1,0 = 0,..., a1,j−2 = 0 and a1,j−1 6= 0. We repeat the last argument so that we obtain a strictly ascending chain of principal ideals of R ⋉n M h(mi )i ( h(a1,i )i ( h(a2,i )i ( · · ·

46


such that, for every k ∈ N, ak,0 = 0,..., ak,j−2 = 0 and ak,j−1 6= 0. Then, by Lemma 7.9, we get a strictly ascending chain of cyclic submodules of Mj−1 hmj−1 i ( ha1,j−1 i ( ha2,j−1 i ( · · · which is absurd by hypothesis on Mj−1 , as desired. Step n + 1. It remains to prove that every element of R ⋉n M of the form (0, ..., 0, mn ) with mn 6= 0 is a product of irreducibles. Since Mn satisfies MCC, Rmn ⊆ Rm where Rm is a maximal cyclic submodule of Mn . Then mn = am for some a ∈ R − {0} and so (0, ..., 0, mn ) = (a, 0, ..., 0)(0, ..., 0, m). Now, a 6= 0 shows that (a, 0, ..., 0) is a product of irreducibles (by Step 1) and Rm is maximal shows that either (0, ..., 0, m) is irreducible or (0, ..., 0, m) = (ai )(bi ) where ak 6= 0 and bl 6= 0 for some k, l ∈ {0, ..., n − 1}. Then by the preceding steps, (ai ) and (bi ) are products of irreducibles and hence so is (0, ..., 0, m). This concludes the proof. A ring S is said to be a bounded factorial ring (BFR) if, for each nonzero nonunit x ∈ S, there is a natural number N (x) so that for any factorization x = x1 · · · xs where each xi is a nonunit, we have s ≤ N (x). For domains we say BFD instead of BFR. Recall that an S-module H is said to be a BF-module if, for each nonzero h ∈ H, there exists a natural number N (h) so that h = a1 · · · as−1 hs (each ai a nonunit) ⇒ s ≤ N (h). Our next result, which is a generalization of [9, Theorem 5.5 (4)], investigates when R ⋉n M is BFR. It is based on the following remark. Lemma 7.12 For j ∈ N − {1}, a product of j elements of R ⋉n M of the form (0, x1 , ..., xn ) is of the form (0, ..., 0, yj , ..., yn ) (where, if j ≥ n + 1 the product is zero). Theorem 7.13 Assume that n ≥ 2, R is an integral domain and Mi is torsion-free for every i ∈ {1, ..., n − 1}. Then R ⋉n M is a BFR if and only if R is a BFD and Mi is a BF-module for every i ∈ {1, ..., n}. Proof. =⇒ Clear. ⇐= Let (mi ) be a nonzero nonunit element of R ⋉n M and suppose we have a factorization into nonunits (mi ) = (a1,i ) · · · (as,i ) for some s ∈ N. If m0 6= 0, m0 = a1,0 · · · as,0 implies that s ≤ N (m0 ). Otherwise, there is j ∈ {1, ..., n} such that m0 = 0,..., mj−1 = 0 and mj 6= 0. We may assume that s ≥ j + 1. Since R is an integral domain and by Lemma 7.12, we may assume there is k ∈ {1, ..., j} such that al,0 = 0 for every l ∈ {1, ..., k} and al,0 6= 0 for every k s Q Q l ∈ {k + 1, ..., s}. Let (0, ..., 0, bk , ..., bn ) = (al,i ) and (c0 , ..., cn ) = (al,i ). Since Mi is l=1

torsion-free for every i ∈ {1, ..., j − 1} and c0 = Then mj = c0 bj =

s Q

s Q

l=k+1

al,0 6= 0, bk = 0,...,bj−1 = 0 and bj 6= 0.

l=k+1

al,0 bj . Therefore s ≤ N (mj ) + k − 1 (since Mj is a BF-module).

l=k+1


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Now we investigate the notion of a U -factorization. It was introduced by Fletcher [35, 36] and developed by Axtell et al. in [11] and [12]. Let S be a ring and consider a nonunit a ∈ S. By a factorization of a we mean a = a1 · · · as where each ai is a nonunit. Recall from [35] that, for a ∈ S, U (a) = {r ∈ S | ∃s ∈ S with rsa = a} = {r ∈ S | r(a) = (a)}. A U -factorization of a is a factorization a = a1 · · · as b1 · · · bt where, for every 1 ≤ i ≤ s, ai ∈ U (b1 · · · bt ) and, for every 1 ≤ i ≤ t, bi ∈ / U (b1 · · · bî · · · bt ). We denote this U -factorization by a = a1 · · · as ⌈b1 · · · bt ⌉ and call a1 ,...,as (resp., b1 ,...,bt ) the irrelevant (resp., the relevant) factors. Our next result investigates when an n-trivial extension is a U -FFR. First, recall the following definitions. A ring S is called a finite factorization ring (FFR) (resp., a U -finite factorization ring (U -FFR)) if every nonzero nonunit of S has only a finite number of factorizations (resp., U -factorizations) up to order and associates (resp., associates on the relevant factors). A ring S is called a weak finite factorization ring (WFFR) (resp., a U -weak finite factorization ring (U -WFFR)) if every nonzero nonunit of R has only a finite number of nonassociate divisors (resp., nonassociate relevant factors). We have FFR ⇒ WFFR and the converse holds in the domain case. But Z2 × Z2 is a WFFR that is not an FFR. However, from [11, Theorem 2.9], U -FFR ⇔ U -WFFR.

The study of the notions above on the classical trivial extensions has lead to consider the following notion (see [11]). Let N be an S-module. For a nonzero element x ∈ N , we say that Sd1 d2 · · · ds x is a reduced submodule factorization if, for every j ∈ {1, ..., s}, dj 6∈ U (S) and for no cancelling and reordering of the dj ’s is it the case that Sd1 d2 · · · ds x = Sd1 d2 · · · dt x where t < s. The module N is said to be a U -FF module if for every nonzero element x ∈ N , there exist only finitely many reduced submodule factorizations Sx = Sd1 d2 · · · dt xk , up to order and associates on the di , as well as up to associates on the xk . In our context, we introduce the following definition. Definition 7.14 Assume n ≥ 2 and consider i ∈ {1, ..., n}. 1. Let mi ∈ Mi −{0}, s ∈ N and (di1 , ..., dis ) ∈ Mi1 ×· · ·×Mis where {i1 , i2 , ..., is } ⊆ {0, ..., n} with i1 + · · · + is = i. We say that Rdi1 di2 · · · dis mi ⊆ Mi is a ϕ-reduced submodule factorization if, for every j ∈ {1, ..., s} such that ij = 0, dij 6∈ U (R) and for no cancelling and reordering of the dj ’s is it the case that Rdi1 di2 · · · dis = Rdi1 di2 · · · dit where t < s. If no ambiguity can arise, a ϕ-reduced submodule factorization is simply called a reduced submodule factorization. 2. The R-module Mi is said to be a ϕ-U -FF module (or simply U -FF module) if, for every nonzero element x ∈ Mi , there exist only finitely many reduced submodule factorizations Rx = Rdi1 di2 · · · dis , up to order and associates on the dij . It is clear that, for i = 1, the notion of U -FF module defined here is the same as the Axtell’s one.

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Based on the proof of [11, Theorem 4.2], it is asserted in [12, Theorem 3.6] that, if R ⋉1 M1 is a U -FFR, then for every nonzero nonunit d ∈ R, there are only finitely many distinct principal ideals h(d, m)i in R ⋉1 M1 . However, a careful reading of this proof shows that the case of ideals h(d, m)i with dM1 = 0 should be also treated. One can confirm the validity of this assertion for reduced rings. However, the context of n-trivial extensions seems to be more complicated. Nevertheless, under some certain conditions, we next investegate when R ⋉n M is a U -FFR. Lemma 7.15 Assume n ≥ 2 and Mn is integral. Then for every nonzero nonunit d ∈ R, the following assertions are true. 1. For every i ∈ {1, ..., n − 1}, the following assertions are equivalent: 1.a. dMi = 0. 1.b. dmi = 0 for some mi ∈ Mi − {0}. 1.c. dMn−i = 0. 1.d. dmn−i = 0 for some mn−i ∈ Mn−i − {0}. 2. The following assertions are equivalent: 2.a. dMi = 0 for some i ∈ {1, ..., n − 1}. 2.b. dMi = 0 for every i ∈ {1, ..., n − 1}. 3. If dMn = 0, then dMi = 0 for every i ∈ {1, ..., n − 1}. 4. If Mn is torsion-free, then Mi is torsion-free for every i ∈ {1, ..., n − 1}. Proof. (1). For the implications (1.a) ⇒ (1.b) and (1.c) ⇒ (1.d) there is nothing to prove. (1.b) ⇒ (1.c). Let m ∈ Mn−i . Then dmi m = 0 ∈ Mn . Therefore dm = 0 (since Mn is integral and mi 6= 0). (1.d) ⇒ (1.a). Similar to the previous proof. (2). For the implication (2.b) ⇒ (2.a) there is nothing to prove. (2.a) ⇒ (2.b). First, we prove that dM1 = 0. For every m1 ∈ M1 − {0}, dmi1 = 0 ∈ Mi and so dmn1 = 0 ∈ Mn . Therefore dm1 = 0 (since Mn is integral and m1 6= 0). Now, consider any j ∈ {1, ..., n − 1} and any mj ∈ Mj − {0}. Then for every m1 ∈ M1 − {0}, dmj m1n−j = 0 ∈ Mn which shows that dmj = 0. (3). This is proved as above. (4). If there is m1 ∈ M1 − {0} and r ∈ R − {0} such that rm1 = 0, then rmn1 = 0 ∈ Mn . Since Mn is torsion-free and r 6= 0, mn1 = 0 ∈ Mn so m1 = 0 (since Mn is integral). This is absurd since m1 6= 0. Finally, by assertions (1) and (2), we conclude that Mi is torsion-free for every i ∈ {1, ..., n − 1}.


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Theorem 7.16 Assume n ≥ 2. If R ⋉n M is a U -FFR (equivalently, a U -WFFR), then the following conditions are satisfied: 1. R is an FFR. 2. Mi is a U -FF module for every i ∈ {1, ..., n}. Moreover, if R is an integral domain and Mn is integral and torsion-free, then 3. For every nonzero nonunit d ∈ R, there are only finitely many distinct principal ideals h(d, m1 , ..., mn )i in R ⋉n M . 4. For every i ∈ {1, ..., n − 1} and every m ∈ Mi − {0}, there are only finitely many distinct principal ideals h(0, ..., 0, m, mi+1 , ..., mn )i in R ⋉n M . Conversely, if R is an integral domain and Mn is integral and torsion-free, then the assertions (1)-(4) imply that R ⋉n M is a U -FFR. Proof. The proof of the “converse” part is similar to the corresponding one of [11, Theorem 4.2]. =⇒ The proof of each (1) and (2) is similar to that given in [11, Theorem 4.2]. 3. Suppose, by contradiction, there exists a nonzero nonunit d ∈ R for which there is a family of distinct principal ideals of the form h(d, mj,1 , ..., mj,n )i where j is in an infinite indexing set Γ. We prove this is impossible by showing that, for every j 6= k in Γ, there exists (1, x1 , ..., xn ) ∈ R ⋉n M such that (d, mj,1 , ..., mj,n ) = (1, x1 , ..., xn )(d, mk,1 , ..., mk,n ). A recursive argument shows that the fact that every equation dX = bi , with bi ∈ Mi admits a solution X ∈ Mi implies the existence of the desired (1, x1 , ..., xn ). Note that, from Lemma 7.15, Mi is torsion-free for every i ∈ {1, ..., n}. First, consider an element bn ∈ Mn − {0}. For every j ∈ Γ, (d, mj,1 , ..., mj,n )(0, ..., 0, bn ) = (0, ..., 0, dbn ). Then (0, ..., 0, dbn ) = (d, mj,1 , ..., mj,n )⌈(0, ..., 0, bn )⌉ is the only possible corresponding U -factorization of (0, ..., 0, dbn ) (since R ⋉n M is a U -FFR), so there exists r ∈ R such that bn = drbn . This shows that the above equation admits a solution for i = n. Now consider k ∈ {1, ..., n−1} and any bk ∈ Mk . For every bn−k ∈ Mn−k − {0}, bk bn−k ∈ Mn − {0} and so there is r ∈ R such that bk bn−k = drbk bn−k . Then (bk − drbk )bn−k = 0. Therefore bk = drbk (since Mn is integral). 4. Let i ∈ {1, ..., n − 1}. Suppose, by contradiction, there exists m ∈ Mi − {0}, for which there is a family of distinct principal ideals of the form h(0, ..., 0, m, mj,i+1 , ..., mj,n )i where j is in an infinite indexing set Γ. Let mn−i ∈ Mn−i − {0}. Necessarily, mmn−i 6= 0. Then (0, ..., 0, m, mj,i+1 , ..., mj,n )(0, ..., 0, mn−i , 0..., 0) = (0, ..., 0, mmn−i ). Then (0, ..., 0, mmn−i ) = (0, ..., 0, m, mj,i+1 , ..., mj,n )⌈(0, ..., 0, mn−i , 0..., 0)⌉ is the only possible corresponding U -factorization of (0, ..., 0, mmn−i ) (since R ⋉n M is a U -FFR), so there exists (r0 , r1 , ..., rn ) such that (0, ..., 0, m, mj,i+1 , ..., mj,n )(r0 , r1 , ..., rn )(0, ..., 0, mn−i , 0..., 0) = (0, ..., 0, mn−i , 0..., 0),

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equivalently (0, ..., 0, r0 mmn−i ) = (0, ..., 0, mn−i , 0..., 0), which is absurd. A ring S is called a U -bounded factorization ring (U -BFR) if, for each nonzero nonunit x ∈ S, there is a natural number N (x) so that, for any factorization x = a⌈b1 · · · bt ⌉, we have t ≤ N (x). An S-module H is said to be a U -BF module if for every h ∈ H − {0} there exists a natural number N (h) so that if Sh = Sd1 · · · dt h′ where dj 6∈ U (S), t > N (h) and h′ ∈ H, then, after cancellation and reordering of some of the dj ’s we have Sh = Sd1 · · · ds h′ for some s ≤ N (h). The question of when the classical trivial extension is a U -BFR is still open. However, there is an answer to this question for an integral domain D [11, Theorem 4.4]: For a D-module N , D ⋉ N is a U -BFR if and only if D is a BFD and N is a U -BF R-module. Two more general results for the direct implication were established in [12, Theorem 3.7 and Lemma 3.8]. Here, we extend these results to our context. For this we need to introduce the following definition.

Definition 7.17 Assume n ≥ 2 and consider i ∈ {1, ..., n}. The R-module Mi is said to be a ϕ-U -BF module (or simply a U -BF module) if, for every nonzero element x ∈ Mi , there exists a natural number N (x) so that if Rx = Rdi1 di2 · · · dit where t ∈ N, (di1 , ..., dit ) ∈ Mi1 × · · · × Mit for some {i1 , i2 , ..., it } ⊆ {0, ..., n} with i1 + · · · + it = i, dij 6∈ U (R) when ij = 0, and t > N (x), then, after cancellation and reordering of some of the dij ’s in R, we have Rx = Rdi1 di2 · · · dis for some s ≤ N (h). Theorem 7.18 If R ⋉n M is a U -BFR, then R is a U -BFR and Mi is a U -BF module for every i ∈ {1, ..., n}. Moreover, if R is présimplifiable, then R is a BFR. Conversely, assume R to be an integral domain. If R is a BFD and for every i ∈ {1, ..., n}, Mi is a U -BF module, then R ⋉n M is a U -BFR. Proof. Similar to the classical case. A ring S is called U -atomic if every nonzero nonunit element of S has a U -factorization in which all the relevant factors are irreducibles. The question of when the classical trivial extension is U -atomic is still unsolved. In [11, Theorem 4.6], Axtell gave an answer to this question for an integral domain D with ACCP: For a D-module N , D ⋉ N is atomic if and only if D ⋉ N is U -atomic. In [12, Theorem 3.15], it is shown that the condition that the ring is an integral domain could be replaced by the ring is présimplifiable. The following result gives an extension of [12, Theorem 3.15] to the context of n-trivial extensions. Theorem 7.19 Assume n ≥ 2. Suppose that Mi is is présimplifiable for every i ∈ {0, ..., n − 1} (here M0 = R), R satisfies ACCP and Mi satisfies ACC on cyclic submodules for every i ∈ {1, ..., n − 1}. Then R ⋉n M is atomic if and only if R ⋉n M is U -atomic. Proof. =⇒ Clear. ⇐= Suppose that R ⋉n M is not atomic. Then by the proof of Theorem 7.11, there exists mn := (0, ..., 0, mn ) ∈ R ⋉n M with mn 6= 0 which is not a product of irreducibles. Since R ⋉n M


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is U -atomic, mn admits a U -factorization of the form mn = a1 · · · as ⌈b1 · · · bt ⌉ such that the bl ’s are irreducibles. Since mn cannot be a product of irreducibles and by the proof of Theorem 7.11, necessarily s = 1 and a1 has the form xn := (0, ..., 0, xn ). But xn hb1 · · · bt i = hb1 · · · bt i and so b1 · · · bt has the form yn := (0, ..., 0, yn ). This is impossible since mn = xn yn = 0.

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